{"id":6035,"date":"2023-08-06T16:31:18","date_gmt":"2023-08-06T16:31:18","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=6035"},"modified":"2023-08-06T16:31:18","modified_gmt":"2023-08-06T16:31:18","slug":"ncert-solutions-class-10-maths-chapter-4-quadratic-equations-ex-4-3","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-4-quadratic-equations-ex-4-3\/","title":{"rendered":"NCERT Solutions Class 10 Maths Chapter 4 Quadratic Equations Ex 4.3"},"content":{"rendered":"<h2 style=\"text-align: center;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">NCERT Solutions Class 10 Maths\u00a0<\/span><br \/>\n<span style=\"color: #000000; font-family: georgia, palatino, serif;\">Chapter &#8211; 4 (Quadratic Equations)\u00a0<\/span><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">The NCERT Solutions in English Language for Class 10 Mathematics <strong>Chapter &#8211; 4 Quadratic Equations <\/strong>Exercise 4.3 has been provided here to help the students in solving the questions from this exercise.\u00a0<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>Chapter : 4 Quadratic Equations<\/strong> <\/span><\/p>\n<ul>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-4-quadratic-equations-ex-4-1\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 4.1<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-4-quadratic-equations-ex-4-2\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 4.2<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-4-quadratic-equations-ex-4-4\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 4.4<\/a><\/span><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 4.3<\/span><\/strong><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>1. Find the roots of the following quadratic equations, if they exist, by the method of completing the square:<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(i) <\/strong>2<em>x<\/em><sup>2<\/sup>\u00a0\u2013 7<em>x<\/em>\u00a0+3 = 0<strong><br \/>\n<\/strong><strong>(ii)\u00a0<\/strong>2<em>x<\/em><sup>2<\/sup>\u00a0+\u00a0<em>x<\/em>\u00a0\u2013 4 = 0<strong><br \/>\n(iii)\u00a0<\/strong>4<em>x<\/em><sup>2<\/sup>\u00a0+ 4\u221a3<em>x<\/em>\u00a0+ 3 = 0<br \/>\n<strong>(iv)\u00a0<\/strong>2<em>x<\/em><sup>2<\/sup>\u00a0+\u00a0<em>x<\/em>\u00a0+ 4 = 0<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solutions &#8211; <\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(i) 2<i>x<\/i><sup>2<\/sup>\u00a0\u2013\u00a07<i>x<\/i> + 3 = 0<br \/>\n<\/strong>\u21d2 2<i>x<\/i><sup>2<\/sup>\u00a0\u2013\u00a07<i>x\u00a0<\/i>= &#8211; 3<\/span><br \/>\n<span style=\"color: #000000;\">Dividing by 2 on both the sides<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 <i>x<\/i><sup>2<\/sup>\u00a0\u2013 7<i>x<\/i>\/2 \u00a0= -3\/2<\/span><br \/>\n<span style=\"color: #000000;\">Since 7\/2 \u00f7 2 = (7\/4), Therefore,\u00a0(7\/4)<sup>2\u00a0<\/sup>\u00a0should be added to both sides of the\u00a0quadratic equation<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 x<sup>2<\/sup>\u00a0&#8211; 7x\/2 + (7\/4)<sup>2<\/sup>\u00a0= &#8211; 3\/2 + (7\/4)<sup>2<br \/>\n<\/sup>\u21d2 (x &#8211; 7\/4)<sup>2<\/sup>\u00a0= &#8211; 3\/2 + (49\/16) [Since, a<sup>2<\/sup>\u00a0&#8211; 2ab + b<sup>2<\/sup>\u00a0= (a &#8211; b)<sup>2<\/sup>]<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 (x &#8211; 7\/4)<sup>2\u00a0<\/sup>= (-24 + 49) \/ 16<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 (x &#8211; 7\/4)<sup>2<\/sup>\u00a0= 25\/16<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 (x &#8211; 7\/4)<sup>2<\/sup>\u00a0= (5\/4)<sup>2<br \/>\n<\/sup>\u21d2 x &#8211; 7\/4 = 5\/4 and x &#8211; 7\/4 = &#8211; 5\/4<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 x = 5\/4 + 7\/4 and x = &#8211; 5\/4 + 7\/4<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 x = 12\/4 and x = 2\/4<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 x = 3 and x = 1\/2<\/span><br \/>\n<span style=\"color: #000000;\">Roots are : 3, 1\/2<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) 2<i>x<\/i><sup>2<\/sup>\u00a0+\u00a0<i>x<\/i>\u00a0\u2013 4 = 0<\/strong><br \/>\n\u21d2 2<i>x<\/i><sup>2<\/sup>\u00a0+\u00a0<i>x\u00a0<\/i>= 4<br \/>\nDividing by 2 on both the sides)<br \/>\n\u21d2\u00a0<i>x<\/i><sup>2<\/sup>\u00a0+\u00a0<i>x<\/i>\/2 = 2<br \/>\nSince 1\/2 \u00f7 2 = 1\/2 \u00d7 1\/2 = 1\/4,<br \/>\n(1\/4)<sup>2<\/sup> should be added to both sides<br \/>\n\u21d2 x<sup>2<\/sup>\u00a0+ x\/2 + (1\/4)<sup>2<\/sup>\u00a0= 2 + (1\/4)<sup>2<br \/>\n<\/sup>\u21d2 [x + (1\/4)]<sup>2<\/sup> = 2 + 1\/16 [Since, a<sup>2<\/sup>\u00a0+\u00a02ab + b<sup>2<\/sup>\u00a0= (a +\u00a0b)<sup>2<\/sup>]<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 [x + (1\/4)]<sup>2<\/sup>\u00a0= (32 + 1) \/ 16<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 [x + (1\/4)]<sup>2<\/sup>\u00a0= 33\/16<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 x + 1\/4 = \u221a33\/4 and x + 1\/4 = &#8211; \u221a33\/4<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 x = \u221a33\/4 &#8211; 1\/4 and x = &#8211; \u221a33\/4 &#8211;\u00a01\/4<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 x = (\u221a33 &#8211; 1)\/4 and x = (- \u221a33 &#8211;\u00a01)\/4<\/span><br \/>\n<span style=\"color: #000000;\">Thus, the roots are\u00a0(\u221a33 &#8211; 1)\/4 and\u00a0(- \u221a33 &#8211;\u00a01)\/4<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii) 4<i>x<\/i><sup>2<\/sup>\u00a0+ 4\u221a3<i>x<\/i>\u00a0+ 3 = 0<\/strong><br \/>\n\u21d2 4<i>x<\/i><sup>2<\/sup>\u00a0+ 4\u221a3<i>x<\/i> = -3<br \/>\nDividing by 4 on both the sides<br \/>\n\u21d2 x<sup>2<\/sup>\u00a0+ \u221a3x = &#8211; 3\/4<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 x<sup>2<\/sup>\u00a0+ \u221a3x + (\u221a3\/2)\u00b2 = &#8211; 3\/4 + (\u221a3\/2)<sup>2<\/sup>\u00a0[(\u221a3\/2)<sup>2<\/sup>\u00a0is added on both sides]<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 [x + (\u221a3\/2)]<sup>2<\/sup>\u00a0= &#8211; 3\/4 + 3\/4\u00a0[Since, a<sup>2<\/sup>\u00a0+\u00a02ab + b<sup>2<\/sup>\u00a0= (a +\u00a0b)<sup>2<\/sup>]<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 [x + (\u221a3\/2)]<sup>2<\/sup>\u00a0= 0<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 x = &#8211; \u221a3\/2 and x = &#8211; \u221a3\/2<\/span><br \/>\n<span style=\"color: #000000;\">Roots\u00a0are &#8211; \u221a3\/2 and &#8211; \u221a3\/2<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iv) 2<i>x<\/i><sup>2<\/sup>\u00a0+\u00a0<i>x<\/i>\u00a0+ 4 = 0<\/strong><br \/>\n\u21d2 2<i>x<\/i><sup>2<\/sup>\u00a0+\u00a0<i>x\u00a0<\/i>= -4<br \/>\nDividing both the sides by 2<br \/>\n\u21d2 <i>x<\/i><sup>2<\/sup> + x\/2 = 2<br \/>\nAdding\u00a0(1\/4)<sup>2\u00a0<\/sup>on both sides of the equation,<br \/>\n\u21d2 x<sup>2<\/sup>\u00a0+ x\/2 + (1\/4)<sup>2<\/sup>\u00a0= &#8211; 2 + (1\/4)<sup>2<br \/>\n<\/sup>\u21d2 [x + (1\/4)]<sup>2<\/sup>\u00a0= &#8211; 2 + 1\/16\u00a0[Since, a<sup>2<\/sup>\u00a0+\u00a02ab + b<sup>2<\/sup>\u00a0= (a +\u00a0b)<sup>2<\/sup>]<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 [x + (1\/4)]<sup>2<\/sup>\u00a0= (- 32 + 1)\/16<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 (x + (1\/4))<sup>2<\/sup>\u00a0= &#8211; 31\/16 &lt; 0<\/span><br \/>\n<span style=\"color: #000000;\">Square\u00a0of any\u00a0real number\u00a0can\u2019t be negative.<\/span><br \/>\n<span style=\"color: #000000;\">Hence, real roots don\u2019t exist.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>2. Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula.<br \/>\n<\/strong><strong>(i) 2x<sup>2<\/sup> \u2013 7x + 3 = 0<br \/>\n<\/strong>On comparing the given equation with\u00a0ax<sup>2<\/sup>\u00a0+\u00a0bx\u00a0+\u00a0c\u00a0= 0, we get,<br \/>\na\u00a0= 2,\u00a0b\u00a0= -7 and\u00a0c\u00a0= 3<br \/>\nBy using the quadratic formula, we get,<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6244 \" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/07\/NCERT-Class-10-Maths-Ex4.3-Ans2.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"214\" height=\"62\" \/><br \/>\n\u21d2 x = (7 \u00b1 \u221a(49 \u2013 24))\/4<br \/>\n\u21d2 x = (7 \u00b1 \u221a25)\/4<br \/>\n\u21d2 x = (7 \u00b1 5)\/4<br \/>\n\u21d2 x = (7 + 5)\/4 or x = (7 &#8211; 5)\/4<br \/>\n\u21d2\u00a0x\u00a0= 12\/4 or 2\/4<br \/>\n\u2234 \u00a0x\u00a0=\u00a03 or 1\/2<br \/>\nThe roots are : 3, \/2<br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) 2x<sup>2<\/sup>\u00a0+\u00a0x\u00a0\u2013 4 = 0<br \/>\n<\/strong>On comparing the given equation with\u00a0ax<sup>2<\/sup>\u00a0+\u00a0bx\u00a0+\u00a0c\u00a0= 0, we get,<br \/>\na = 2, b = 1 and c = -4<br \/>\nBy using the quadratic formula, we get,<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6244 \" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/07\/NCERT-Class-10-Maths-Ex4.3-Ans2.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"214\" height=\"62\" \/><br \/>\n\u21d2 x = (-1 \u00b1 \u221a1 + 32)\/4<br \/>\n\u21d2 x = (-1 \u00b1 \u221a33)\/4<br \/>\n\u2234 x = (-1 + \u221a33)\/4 or x = (-1 &#8211; \u221a33)\/4<br \/>\nThe roots are : (-1 + \u221a33)\/4, (-1 &#8211; \u221a33)\/4<br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii) 4x<sup>2<\/sup>\u00a0+\u00a04\u221a3x\u00a0+ 3 = 0<br \/>\n<\/strong>On comparing the given equation with\u00a0ax<sup>2<\/sup>\u00a0+\u00a0bx\u00a0+\u00a0c\u00a0= 0, we get,<br \/>\na\u00a0=\u00a04,\u00a0b\u00a0=\u00a04\u221a3\u00a0and\u00a0c = 3<br \/>\nBy using the quadratic formula, we get,<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6244 \" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/07\/NCERT-Class-10-Maths-Ex4.3-Ans2.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"214\" height=\"62\" \/><br \/>\n\u21d2 x = (-4\u221a3 \u00b1 \u221a48 &#8211; 48)\/8<br \/>\n\u21d2 x = (-4\u221a3 \u00b1 0)\/8<br \/>\n\u2234\u00a0x\u00a0= -\u221a3\/2 or\u00a0x\u00a0= -\u221a3\/2<br \/>\nThe roots are : -\u221a3\/2, -\u221a3\/2<br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iv)\u00a02x<sup>2<\/sup>\u00a0+\u00a0x\u00a0+ 4 = 0<br \/>\n<\/strong>On comparing the given equation with\u00a0ax<sup>2<\/sup>\u00a0+\u00a0bx\u00a0+\u00a0c\u00a0= 0, we get,<br \/>\na\u00a0= 2,\u00a0b\u00a0= 1 and\u00a0c\u00a0= 4<br \/>\nBy using the quadratic formula, we get<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6244 \" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/07\/NCERT-Class-10-Maths-Ex4.3-Ans2.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"214\" height=\"62\" \/><br \/>\n\u21d2 x = (-1 \u00b1 \u221a1 &#8211; 32)\/4<br \/>\n\u21d2 x = (-1 \u00b1 \u221a-31)\/4<br \/>\nAs we know, the square of a number can never be negative. Therefore, there is no real solution for the given equation.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>3. Find the roots of the following equations:<br \/>\n<\/strong><strong>(i) x &#8211; <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{1}{x}}\" alt=\"\\mathbf{\\frac{1}{x}}\" align=\"absmiddle\" \/> = 3, x \u2260 0<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{1}{x+4}}\" alt=\"\\mathbf{\\frac{1}{x+4}}\" align=\"absmiddle\" \/> \u2013 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{1}{x-7}}\" alt=\"\\mathbf{\\frac{1}{x-7}}\" align=\"absmiddle\" \/> = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{11}{30}}\" alt=\"\\mathbf{\\frac{11}{30}}\" align=\"absmiddle\" \/>, x = -4, 7<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution:<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(i) x &#8211; <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{1}{x}}\" alt=\"\\mathbf{\\frac{1}{x}}\" align=\"absmiddle\" \/> = 3<br \/>\n<\/strong>Multiplying x on both the sides<strong><br \/>\n<\/strong>\u21d2 x<sup>2<\/sup> &#8211; 1 = 3x<br \/>\n\u21d2 x<sup>2<\/sup> &#8211; 3x &#8211; 1 = 0<br \/>\nOn comparing the given equation with\u00a0ax<sup>2<\/sup>\u00a0+\u00a0bx\u00a0+\u00a0c\u00a0= 0, we get<br \/>\na\u00a0= 1,\u00a0b\u00a0= -3 and\u00a0c\u00a0= -1<br \/>\nBy using the quadratic formula, we get<br \/>\n<img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6244 \" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/07\/NCERT-Class-10-Maths-Ex4.3-Ans2.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"214\" height=\"62\" \/><br \/>\n\u21d2 x = (3 \u00b1 \u221a9 + 4)\/2<br \/>\n\u21d2 x = (3 \u00b1 \u221a13)\/2<br \/>\n\u2234 x = (3 + \u221a13)\/2 or x = (3 &#8211; \u221a13)\/2<br \/>\nThe roots are : (3 + \u221a13) \/ 2, (3 &#8211; \u221a13) \/ 2<br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) 1\/x+4 \u2013 1\/x-7 = 11\/30<br \/>\n<\/strong>\u21d2 [(x &#8211; 7) &#8211; (x + 4)] \/ (x + 4)(x &#8211; 7) = 11\/ 30<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 [x &#8211; 7 &#8211; x &#8211; 4] \/ x<sup>2<\/sup>\u00a0+ 4x &#8211; 7x &#8211; 28 = 11\/ 30<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 (- 11) \/ (x<sup>2<\/sup>\u00a0&#8211; 3x &#8211; 28) = 11\/30<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 &#8211; 11 \u00d7 30 = 11(x<sup>2<\/sup>\u00a0&#8211; 3x &#8211; 28)<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 &#8211; 30 = (x<sup>2<\/sup> &#8211; 3x &#8211; 28) [Cancelling 11 from both sides of the equation]<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 x<sup>2<\/sup>\u00a0&#8211; 3x &#8211; 28 + 30 = 0<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 x<sup>2<\/sup>\u00a0&#8211; 3x + 2 = 0<\/span><br \/>\n<span style=\"color: #000000;\">Comparing x<sup>2<\/sup>\u00a0&#8211; 3x + 2 = 0 against the standard form ax<sup>2<\/sup>\u00a0&#8211; bx + c = 0 ,<\/span><br \/>\n<span style=\"color: #000000;\">We find that a = 1, b = &#8211; 3 and c = 2<\/span><br \/>\n<span style=\"color: #000000;\">b<sup>2<\/sup>\u00a0&#8211; 4ac = (-3)<sup>2<\/sup>\u00a0&#8211; 4(1)(2)<\/span><br \/>\n<span style=\"color: #000000;\">= 9 &#8211; 8<\/span><br \/>\n<span style=\"color: #000000;\">= 1 &gt; 0<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, real roots exist for this quadratic equation.<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, <\/span><br \/>\n<span style=\"color: #000000;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6244 \" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/07\/NCERT-Class-10-Maths-Ex4.3-Ans2.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"214\" height=\"62\" \/><br \/>\nx =\u00a0(3 \u00b1 \u221a1) \/ 2<\/span><br \/>\n<span style=\"color: #000000;\">x = (3 + 1)\/2 and x = (3 &#8211; 1)\/2<\/span><br \/>\n<span style=\"color: #000000;\">x = 4 \/ 2 and x = 2 \/ 2<\/span><br \/>\n<span style=\"color: #000000;\">x = 2 and x = 1<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, roots are 2, 1.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>4. The sum of the reciprocals of Rehman\u2019s ages (in years) 3 years ago and 5 years from now is 1\/3. Find his present age.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;\u00a0 <\/strong>Let the present age of Rehman be\u00a0<i>x<\/i>\u00a0years.<\/span><br \/>\n<span style=\"color: #000000;\">Three years ago, his age was (<i>x<\/i>\u00a0&#8211; 3) years.<\/span><br \/>\n<span style=\"color: #000000;\">Five years hence, his age will be (<i>x<\/i>\u00a0+ 5) years.<\/span><br \/>\n<span style=\"color: #000000;\">It is given that the sum of the reciprocals of Rehman&#8217;s ages 3 years ago and 5 years from now is 1\/3.<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{x-3}\" alt=\"\\frac{1}{x-3}\" align=\"absmiddle\" \/> + <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{x-5}\" alt=\"\\frac{1}{x-5}\" align=\"absmiddle\" \/> = <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{1}{3}\" alt=\"\\frac{1}{3}\" align=\"absmiddle\" \/><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">\u21d2 [(x + 5) + (x &#8211; 3)] \/ (x &#8211; 3)(x + 5) = 1\/3<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 (2x + 2) \/ x<sup>2<\/sup>\u00a0+ 2x &#8211; 15 = 1\/3<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 (2x + 2)(3) = x<sup>2<\/sup>\u00a0+ 2x &#8211; 15<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 6x + 6 = x<sup>2<\/sup>\u00a0+ 2x &#8211; 15<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 x<sup>2<\/sup>\u00a0+ 2x &#8211; 15 = 6x + 6<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 x<sup>2<\/sup>\u00a0&#8211; 4x &#8211; 21 = 0<\/span><br \/>\n<span style=\"color: #000000;\">Let us now find the\u00a0roots by\u00a0factorization\u00a0method:<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 x<sup>2<\/sup>\u00a0&#8211; 7x + 3x &#8211; 21 = 0<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 x(x &#8211; 7) + 3(x &#8211; 7) = 0<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 (x &#8211; 7) (x + 3) = 0<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 x &#8211; 7 = 0 and x + 3 = 0<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 x = 7 and x = &#8211; 3<\/span><br \/>\n<span style=\"color: #000000;\">But, age can\u2019t be a negative value.<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, Rehman\u2019s present age is 7 years.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>5. In a class test, the sum of Shefali\u2019s marks in Mathematics and English is 30. If she had got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;\u00a0 <\/strong>Let the marks in Maths be\u00a0<i>x<\/i>.<br \/>\nThen, the marks in English will be 30 &#8211;\u00a0<i>x<\/i>.<br \/>\nAccording to the question,<br \/>\n(<i>x<\/i>\u00a0+ 2)(30 &#8211;\u00a0<i>x<\/i>\u00a0&#8211; 3) = 210<br \/>\n(<i>x<\/i>\u00a0+ 2)(27 &#8211;\u00a0<i>x<\/i>) = 210<br \/>\n\u21d2 &#8211;<i>x<\/i><sup>2<\/sup>\u00a0+ 25<i>x<\/i>\u00a0+ 54 = 210<br \/>\n\u21d2\u00a0<i>x<\/i><sup>2<\/sup>\u00a0&#8211; 25<i>x<\/i>\u00a0+ 156 = 0<br \/>\n\u21d2\u00a0<i>x<\/i><sup>2<\/sup>\u00a0&#8211;\u00a012<i>x<\/i>\u00a0&#8211; 13<i>x<\/i>\u00a0+ 156 = 0<br \/>\n\u21d2\u00a0<i>x<\/i>(<i>x<\/i>\u00a0&#8211; 12) -13(<i>x<\/i>\u00a0&#8211; 12) = 0<br \/>\n\u21d2 (<i>x<\/i>\u00a0&#8211; 12)(<i>x<\/i>\u00a0&#8211; 13) = 0<br \/>\n\u21d2\u00a0<i>x<\/i>\u00a0= 12, 13<br \/>\nIf the marks in Maths are 12, then marks in English will be 30 &#8211; 12 = 18<br \/>\nIf the marks in Maths are 13, then marks in English will be 30 &#8211; 13 = 17<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>6.\u00a0The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;\u00a0 <\/strong>Let the shorter side of the rectangle be\u00a0<i>x<\/i> m.<br \/>\nThen, larger side of the rectangle = (<i>x<\/i>\u00a0+ 30) m<br \/>\nBy applying\u00a0Pythagoras theorem:<\/span><br \/>\n<span style=\"color: #000000;\">Hypotenuse\u00b2\u00a0= Side 1<sup>2<\/sup>\u00a0+ Side 2<sup>2<br \/>\n<\/sup>(60 + x)<sup>2<\/sup>\u00a0= x<sup>2<\/sup>\u00a0+ (30 + x)<sup>2<br \/>\n<\/sup>60<sup>2<\/sup>\u00a0+ 2(60)x + x<sup>2<\/sup>\u00a0= x<sup>2<\/sup>\u00a0+ 30<sup>2<\/sup>\u00a0+ 2(30)x + x<sup>2<br \/>\n<\/sup>3600 + 120x + x<sup>2<\/sup>\u00a0= x<sup>2<\/sup>\u00a0+ 900 + 60x + x<sup>2<br \/>\n<\/sup>3600 + 120x + x<sup>2<\/sup>\u00a0&#8211; x<sup>2<\/sup>\u00a0&#8211; 900 &#8211; 60x &#8211; x<sup>2<\/sup>\u00a0= 0<\/span><br \/>\n<span style=\"color: #000000;\">2700 + 60x &#8211; x<sup>2<\/sup>\u00a0= 0<\/span><br \/>\n<span style=\"color: #000000;\">Multiplying both sides by -1:<\/span><br \/>\n<span style=\"color: #000000;\">x<sup>2<\/sup>\u00a0&#8211; 60x &#8211; 2700 = 0<\/span><br \/>\n<span style=\"color: #000000;\">Comparing x<sup>2<\/sup>\u00a0&#8211; 60x &#8211; 2700 = 0 with ax<sup>2<\/sup>\u00a0+ bx + c = 0 we get,<\/span><br \/>\n<span style=\"color: #000000;\">a = 1, b = &#8211; 60, c = &#8211; 2700<\/span><br \/>\n<span style=\"color: #000000;\">b<sup>2<\/sup>\u00a0&#8211; 4ac = (-60)<sup>2<\/sup>\u00a0&#8211; 4(1)(-2700)<\/span><br \/>\n<span style=\"color: #000000;\">b<sup>2<\/sup>\u00a0&#8211; 4ac = 3600 + 10800<\/span><br \/>\n<span style=\"color: #000000;\">b<sup>2<\/sup>\u00a0&#8211; 4ac = 14400 &gt; 0<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 Roots exist.<\/span><br \/>\n<span style=\"color: #000000;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6244 \" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/07\/NCERT-Class-10-Maths-Ex4.3-Ans2.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"214\" height=\"62\" \/><\/span><br \/>\n<span style=\"color: #000000;\">= [-(- 60)\u00a0\u00b1 \u221a(14400)]\u00a0\/ 2<\/span><br \/>\n<span style=\"color: #000000;\">= [(60) \u00b1\u00a0120] \/ 2<\/span><br \/>\n<span style=\"color: #000000;\">x = (60 + 120) \/ 2 and x = (60 &#8211; 120) \/ 2<\/span><br \/>\n<span style=\"color: #000000;\">x = 180 \/ 2 and x = &#8211; 60 \/ 2<\/span><br \/>\n<span style=\"color: #000000;\">x = 90 and a = &#8211; 30<\/span><br \/>\n<span style=\"color: #000000;\">Length can\u2019t be a negative value.<\/span><br \/>\n<span style=\"color: #000000;\">Hence, x = 90<\/span><br \/>\n<span style=\"color: #000000;\">Length of shorter side is x = 90 m<\/span><br \/>\n<span style=\"color: #000000;\">Length of longer side = 30 + x = 30 + 90 = 120 m<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>7. The difference between the squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;\u00a0 <\/strong>Let the larger and smaller number be\u00a0<i>x<\/i>\u00a0and\u00a0<i>y<\/i>\u00a0respectively.<br \/>\nAccording to the question,<br \/>\n<i>x<\/i><sup>2\u00a0<\/sup>&#8211;\u00a0<i>y<\/i><sup>2<\/sup>\u00a0= 180 and\u00a0<i>y<\/i><sup>2<\/sup>\u00a0= 8<i>x<\/i><br \/>\n\u21d2\u00a0<i>x<\/i><sup>2\u00a0<\/sup>&#8211; 8<i>x<\/i>\u00a0= 180<br \/>\n\u21d2\u00a0<i>x<\/i><sup>2\u00a0<\/sup>&#8211;\u00a08<i>x<\/i>\u00a0&#8211; 180 = 0<br \/>\n\u21d2\u00a0<i>x<\/i><sup>2\u00a0<\/sup>&#8211; 18<i>x<\/i>\u00a0+ 10<i>x<\/i>\u00a0&#8211; 180 = 0<br \/>\n\u21d2\u00a0<i>x<\/i>(<i>x<\/i>\u00a0&#8211; 18)\u00a0+10(<i>x<\/i>\u00a0&#8211; 18) = 0<br \/>\n\u21d2 (<i>x<\/i>\u00a0&#8211; 18)(<i>x<\/i>\u00a0+ 10) = 0<br \/>\n\u21d2\u00a0<i>x<\/i> = 18, -10<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Case 1:<\/strong>\u00a0If the larger number is 18, then\u00a0square of smaller number = 8 \u00d7 18 <\/span><br \/>\n<span style=\"color: #000000;\">Therefore, smaller number = \u00b1 \u221a8 \u00d7 18<\/span><br \/>\n<span style=\"color: #000000;\">= \u00b1 \u221a (<u>2 \u00d7 2 \u00d7 2<\/u>\u00a0\u00d7\u00a0<u>3 \u00d7 3 \u00d7 2<\/u>)<\/span><br \/>\n<span style=\"color: #000000;\">= \u00b1 (2 \u00d7 2 \u00d7 3)<\/span><br \/>\n<span style=\"color: #000000;\">= \u00b1 12<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>Case 2:<\/strong>\u00a0If larger number is &#8211; 10, then square of smaller number = 8 \u00d7 (- 10) = &#8211; 80<\/span><br \/>\n<span style=\"color: #000000;\">The square of any number cannot be negative.<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 x = -10 is not applicable.<\/span><br \/>\n<span style=\"color: #000000;\">Thus, the numbers are 18, 12 (or) 18, &#8211; 12.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>8. A train travels 360 km at a uniform speed. If the speed had been 5 km\/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;\u00a0<\/strong>Let the speed of the train be\u00a0<i>x<\/i>\u00a0km\/hr.<br \/>\nDistance = Speed \u00d7 Time<br \/>\n360 = <em>x<\/em> \u00d7 t<br \/>\n\u21d2 t = 360 \/ <em>x<br \/>\n<\/em>Increased speed of the train can be written as <em>x<\/em> + 5<\/span><br \/>\n<span style=\"color: #000000;\">New time to cover the same distance = t &#8211; 1<\/span><br \/>\n<span style=\"color: #000000;\">(x + 5) \u00d7 (t &#8211; 1) = 360\u00a0 \u00a0&#8212;&#8212;&#8212;&#8212; (i)<\/span><br \/>\n<span style=\"color: #000000;\">xt &#8211; x + 5t &#8211; 5 = 360<\/span><br \/>\n<span style=\"color: #000000;\">360 &#8211; x + 5(360\/x) &#8211; 5 = 360 [Since, xt = 360 and t = 360 \/ x]<\/span><br \/>\n<span style=\"color: #000000;\">&#8211; x + 1800\/x &#8211; 5 = 0<\/span><br \/>\n<span style=\"color: #000000;\">&#8211; x\u00b2 + 1800 &#8211; 5x = 0<\/span><br \/>\n<span style=\"color: #000000;\">x\u00b2 + 5x &#8211; 1800 = 0<\/span><br \/>\n<span style=\"color: #000000;\">Comparing x\u00b2 + 5x &#8211; 1800 = 0 with ax<sup>2<\/sup>\u00a0+ bx + c = 0 we get,<\/span><br \/>\n<span style=\"color: #000000;\">a = 1, b = 5, c = &#8211; 1800<\/span><br \/>\n<span style=\"color: #000000;\">b\u00b2 &#8211; 4ac = (5)<sup>2<\/sup>\u00a0&#8211; 4(1)(- 1800)<\/span><br \/>\n<span style=\"color: #000000;\">b\u00b2 &#8211; 4ac = 25 + 7200<\/span><br \/>\n<span style=\"color: #000000;\">b\u00b2 &#8211; 4ac = 7225 &gt; 0<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 Roots exist.<\/span><br \/>\n<span style=\"color: #000000;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-6244 \" src=\"https:\/\/theexampillar.com\/ncert\/wp-content\/uploads\/2023\/07\/NCERT-Class-10-Maths-Ex4.3-Ans2.png\" alt=\"NCERT Class 10 Maths Solution\" width=\"214\" height=\"62\" \/><br \/>\nx = (- 5 \u00b1 \u221a7225) \/ 2<\/span><br \/>\n<span style=\"color: #000000;\">x = (- 5 \u00b1 85) \/ 2<\/span><br \/>\n<span style=\"color: #000000;\">x = (- 5 + 85) \/ 2 and x = (- 5 &#8211; 85) \/ 2<\/span><br \/>\n<span style=\"color: #000000;\">x = 80 \/ 2 and x = &#8211; 90 \/ 2<\/span><br \/>\n<span style=\"color: #000000;\">x = 40 and x = &#8211; 45<\/span><br \/>\n<span style=\"color: #000000;\">Speed of the train cannot be a negative value.<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, speed of the train is 40 km \/hr.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>9. Two water taps together can fill a tank in <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{9\\frac{3}{8}}\" alt=\"\\mathbf{9\\frac{3}{8}}\" align=\"absmiddle\" \/><\/strong><strong>\u00a0hours. The tap of the larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time at which each tap can separately fill the tank.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211; <\/strong>Let the time taken by the smaller pipe to fill the tank be x hr.<br \/>\nTime taken by the larger pipe = (<i>x<\/i> &#8211; 10) hr<br \/>\nPart of tank filled by smaller pipe in 1 hour = 1\/<i>x<br \/>\n<\/i>Part of the tank filled by larger pipe in 1 hour = 1\/(<em>x\u00a0<\/em>\u2013 10)<br \/>\nAs given, the tank can be filled in <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?9\\frac{3}{8}\" alt=\"9\\frac{3}{8}\" align=\"absmiddle\" \/>\u00a0= 75\/8\u00a0hours by both the pipes together.<br \/>\nTherefore,<br \/>\n1\/<em>x<\/em> + 1\/(<em>x <\/em>&#8211; 10) = 8\/75<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{(x-10)+x}{x(x-10)}\" alt=\"\\frac{(x-10)+x}{x(x-10)}\" align=\"absmiddle\" \/> = 8\/75<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">\u21d2 <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\frac{2x&amp;space;-&amp;space;10}{x(x-10)}\" alt=\"\\frac{2x - 10}{x(x-10)}\" align=\"absmiddle\" \/> = 8\/75<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\">\u21d2 75(2<em>x<\/em>\u00a0\u2013 10) = 8<em>x<\/em><sup>2<\/sup>\u00a0\u2013 80<em>x<br \/>\n<\/em>\u21d2 150<em>x<\/em>\u00a0\u2013 750 = 8<em>x<\/em><sup>2<\/sup>\u00a0\u2013 80<em>x<br \/>\n<\/em>\u21d2 8<em>x<\/em><sup>2<\/sup>\u00a0\u2013 230<em>x<\/em>\u00a0+750 = 0<br \/>\n\u21d2 8<em>x<\/em><sup>2<\/sup>\u00a0\u2013 200<em>x<\/em>\u00a0\u2013 30<em>x<\/em>\u00a0+ 750 = 0<br \/>\n\u21d2 8<em>x<\/em>(<em>x<\/em>\u00a0\u2013 25) -30(<em>x<\/em>\u00a0\u2013 25) = 0<br \/>\n\u21d2 (<em>x<\/em>\u00a0\u2013 25)(8<em>x<\/em>\u00a0-30) = 0<br \/>\n\u21d2\u00a0<em>x<\/em>\u00a0= 25, 30\/8<br \/>\nTime taken by the smaller pipe cannot be 30\/8 = 3.75 hours, as the time taken by the larger pipe will become negative, which is logically not possible.<br \/>\nTherefore, the time taken individually by the smaller pipe and the larger pipe will be 25 and 25 \u2013 10 =15 hours, respectively.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>10. An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km\/h more than that of the passenger train, find the average speed of the two trains.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;\u00a0<\/strong>Let the average speed of passenger train be\u00a0<i>x<\/i>\u00a0km\/h.<br \/>\nAverage speed of express train = (<i>x<\/i>\u00a0+ 11) km\/h<br \/>\nIt is given that the time taken by the express train to cover 132 km is 1 hour less than the passenger train to cover the same distance.<br \/>\n(132\/x) \u2013 (132\/(x + 11)) = 1<br \/>\n132(x + 11 &#8211; x)\/(x(x + 11)) = 1<br \/>\n132 \u00d7 11 \/ (x(x + 11)) = 1<br \/>\n\u21d2 132 \u00d7 11 =\u00a0<em>x<\/em>(<em>x<\/em>\u00a0+ 11)<br \/>\n\u21d2\u00a0<em>x<\/em><sup>2<\/sup>\u00a0+ 11<em>x<\/em>\u00a0\u2013 1452 = 0<br \/>\n\u21d2\u00a0<em>x<\/em><sup>2<\/sup>\u00a0+\u00a0 44<em>x<\/em>\u00a0-33<em>x<\/em>\u00a0-1452 = 0<br \/>\n\u21d2\u00a0<em>x<\/em>(<em>x<\/em>\u00a0+ 44) -33(<em>x<\/em>\u00a0+ 44) = 0<br \/>\n\u21d2 (<em>x<\/em>\u00a0+ 44)(<em>x<\/em>\u00a0\u2013 33) = 0<br \/>\n\u21d2\u00a0<em>x<\/em>\u00a0= \u2013 44, 33<br \/>\nAs we know, speed cannot be negative.<br \/>\nTherefore, the speed of the passenger train will be 33 km\/h and thus, the speed of the express train will be 33 + 11 = 44 km\/h.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>11. Sum of the areas of two squares is 468 m<sup>2<\/sup>. If the difference of their perimeters is 24 m, find the sides of the two squares.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;\u00a0 <\/strong>Let the sides of the two squares be\u00a0<em>x<\/em>\u00a0m and\u00a0<em>y<\/em>\u00a0m.<br \/>\nTherefore, their perimeter will be 4<em>x<\/em>\u00a0and 4<em>y,<\/em>\u00a0respectively<br \/>\nAnd the area of the squares will be\u00a0<em>x<\/em><sup>2<\/sup>\u00a0and\u00a0<em>y<\/em><sup>2,\u00a0<\/sup>respectively.<br \/>\nGiven,<br \/>\n4<em>x<\/em>\u00a0\u2013 4<em>y<\/em> = 24<br \/>\n<em>x<\/em>\u00a0\u2013\u00a0<em>y<\/em> = 6<br \/>\n<em>x<\/em>\u00a0=\u00a0<em>y<\/em> + 6<br \/>\nAlso,\u00a0<em>x<\/em><sup>2\u00a0<\/sup>+<sup>\u00a0<\/sup><em>y<\/em><sup>2<\/sup> = 468<br \/>\n\u21d2 (6\u00a0+\u00a0<em>y<\/em><sup>2<\/sup>)\u00a0+\u00a0<em>y<\/em><sup>2<\/sup> = 468<br \/>\n\u21d2 36\u00a0+\u00a0<em>y<\/em><sup>2<\/sup>\u00a0+ 12<em>y<\/em>\u00a0+\u00a0<em>y<\/em><sup>2<\/sup>\u00a0= 468<br \/>\n\u21d2 2<em>y<\/em><sup>2<\/sup>\u00a0+ 12<em>y<\/em>\u00a0+ 432 = 0<br \/>\n\u21d2\u00a0<em>y<\/em><sup>2<\/sup>\u00a0+ 6y \u2013 216 = 0<br \/>\n\u21d2\u00a0<em>y<\/em><sup>2<\/sup>\u00a0+ 18<em>y<\/em>\u00a0\u2013 12<em>y<\/em>\u00a0\u2013 216 = 0<br \/>\n\u21d2\u00a0<em>y<\/em>(<em>y\u00a0<\/em>+18) -12(<em>y<\/em>\u00a0+ 18) = 0<br \/>\n\u21d2 (<em>y<\/em>\u00a0+ 18)(<em>y<\/em>\u00a0\u2013 12) = 0<br \/>\n\u21d2\u00a0<em>y<\/em>\u00a0= -18, 12<br \/>\nAs we know, the side of a square cannot be negative.<br \/>\nHence, the sides of the squares are 12 m and (12 + 6) m = 18 m.<\/span><\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"font-family: Georgia, Palatino;\"><a href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths\/\"><span style=\"color: #0000ff;\"><b><i>Go Back To Chapters<\/i><\/b><\/span><\/a><\/span><\/p>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 10 Maths\u00a0 Chapter &#8211; 4 (Quadratic Equations)\u00a0 The NCERT Solutions in English Language for Class 10 Mathematics Chapter &#8211; 4 Quadratic Equations Exercise 4.3 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter : 4 Quadratic Equations NCERT Class 10 Maths Solution Ex &#8211; 4.1 [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1043],"tags":[1045,1069,1072,1044,1049,1048],"class_list":["post-6035","post","type-post","status-publish","format-standard","hentry","category-class-10-maths","tag-class-10-ncert-mathematics-solutions","tag-ncert-class-10-mathematics-chapter-4-quadratic-equations-solutions","tag-ncert-class-10-mathematics-exercise-4-3-solutions","tag-ncert-class-10-mathematics-solutions-class-10-ncert-solutions","tag-ncert-solutions-class-10-mathematics","tag-ncert-solutions-class-10-maths"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.9 (Yoast SEO v27.3) - 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