{"id":6036,"date":"2023-08-06T16:31:31","date_gmt":"2023-08-06T16:31:31","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=6036"},"modified":"2023-08-06T16:31:31","modified_gmt":"2023-08-06T16:31:31","slug":"ncert-solutions-class-10-maths-chapter-4-quadratic-equations-ex-4-4","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-4-quadratic-equations-ex-4-4\/","title":{"rendered":"NCERT Solutions Class 10 Maths Chapter 4 Quadratic Equations Ex 4.4"},"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 4.4 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-4-3\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 4.3<\/a><\/span><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 4.4<\/span><\/strong><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:<br \/>\n(i) <\/strong>2<em>x<\/em><sup>2<\/sup>\u00a0\u2013 3<em>x<\/em>\u00a0+ 5 = 0<strong><br \/>\n(ii) <\/strong>3<em>x<\/em><sup>2<\/sup>\u00a0\u2013\u00a04\u221a3<em>x<\/em>\u00a0+ 4 = 0<strong><br \/>\n(iii) <\/strong>2<em>x<\/em><sup>2<\/sup>\u00a0\u2013\u00a06<em>x<\/em>\u00a0+ 3 = 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<em>x<\/em><sup>2<\/sup>\u00a0\u2013 3<em>x<\/em>\u00a0+ 5 = 0<br \/>\n<\/strong>Comparing the equation with\u00a0<em>ax<\/em><sup>2<\/sup>\u00a0+\u00a0<em>bx\u00a0<\/em>+\u00a0<em>c<\/em> = 0, we get<br \/>\n<em>a<\/em>\u00a0= 2,\u00a0<em>b<\/em>\u00a0= -3 and\u00a0<em>c<\/em>\u00a0= 5<br \/>\nWe know, discriminant =\u00a0<em>b<\/em><sup>2<\/sup>\u00a0\u2013 4<em>ac<br \/>\n<\/em>\u21d2 ( \u2013 3)<sup>2<\/sup>\u00a0\u2013 4 (2) (5)<br \/>\n\u21d2 9 \u2013 40<br \/>\n\u21d2 \u2013 31<br \/>\nAs you can see, b<sup>2<\/sup> \u2013 4ac &lt; 0<br \/>\nTherefore, no real root is possible for the given equation,\u00a0<em>2x<\/em><sup>2<\/sup>\u00a0\u2013 3<em>x<\/em>\u00a0+ 5 = 0.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) 3<em>x<\/em><sup>2<\/sup>\u00a0\u2013 4\u221a3<em>x<\/em>\u00a0+ 4 = 0<br \/>\n<\/strong>Comparing the equation with\u00a0<em>ax<\/em><sup>2<\/sup>\u00a0+\u00a0<em>bx\u00a0<\/em>+\u00a0<em>c<\/em>\u00a0= 0, we get<br \/>\n<em>a<\/em>\u00a0= 3,\u00a0<em>b<\/em>\u00a0=\u00a0-4\u221a3\u00a0and\u00a0<em>c<\/em> = 4<br \/>\nWe know, Discriminant =\u00a0<em>b<\/em><sup>2<\/sup>\u00a0\u2013 4<em>ac<br \/>\n<\/em>\u21d2 (-4\u221a3)<sup>2\u00a0<\/sup>\u2013 4(3)(4)<br \/>\n\u21d2 48 \u2013 48 = 0<br \/>\nAs\u00a0<em>b<\/em><sup>2<\/sup>\u00a0\u2013 4<em>ac<\/em> = 0,<br \/>\nReal roots exist for the given equation, and they are equal to each other.<br \/>\nHence, the roots will be \u2013<em>b<\/em>\/2<em>a<\/em>\u00a0and\u00a0\u2013<em>b<\/em>\/2<em>a<\/em>.<br \/>\n\u2013<em>b<\/em>\/2<em>a\u00a0<\/em>= -(-4\u221a3)\/2\u00d73 = 4\u221a3\/6 = 2\u221a3\/3 = 2\/\u221a3<br \/>\nTherefore, the roots are\u00a02\/\u221a3\u00a0and 2\/\u221a3.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii) 2<em>x<\/em><sup>2<\/sup>\u00a0\u2013\u00a06<em>x<\/em>\u00a0+ 3 = 0<br \/>\n<\/strong>Comparing the equation with\u00a0<em>ax<\/em><sup>2<\/sup>\u00a0+\u00a0<em>bx\u00a0<\/em>+\u00a0<em>c<\/em>\u00a0= 0, we get<br \/>\n<em>a<\/em>\u00a0= 2,\u00a0<em>b<\/em>\u00a0= -6,\u00a0<em>c<\/em>\u00a0= 3<br \/>\nAs we know, discriminant =\u00a0<em>b<\/em><sup>2<\/sup>\u00a0\u2013 4<em>ac<br \/>\n<\/em>\u21d2 (-6)<sup>2<\/sup>\u00a0\u2013 4 (2) (3)<br \/>\n\u21d2 36 \u2013 24<br \/>\n= 12<br \/>\nAs\u00a0<em>b<\/em><sup>2<\/sup>\u00a0\u2013 4<em>ac<\/em>\u00a0&gt; 0,<br \/>\nTherefore, there are distinct real roots that exist for this equation, 2<em>x<\/em><sup>2<\/sup>\u00a0\u2013\u00a06<em>x<\/em>\u00a0+ 3 = 0.<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=\"211\" height=\"61\" \/><br \/>\n\u21d2 (-(-6) \u00b1 \u221a(-6<sup>2 <\/sup>&#8211; 4(2)(3)))\/ 2(2)<br \/>\n\u21d2 (6\u00b12\u221a3 )\/4<br \/>\n\u21d2 (3\u00b1\u221a3)\/2<br \/>\nTherefore, the roots for the given equation are (3+\u221a3)\/2 and (3-\u221a3)\/2.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>2. Find the values of\u00a0<em>k<\/em>\u00a0for each of the following quadratic equations so that they have two equal roots.<br \/>\n(i) <\/strong>2<em>x<\/em><sup>2<\/sup>\u00a0+\u00a0<em>kx<\/em>\u00a0+ 3 = 0<br \/>\n<strong>(ii)\u00a0<\/strong><em>kx<\/em>\u00a0(<em>x<\/em>\u00a0\u2013 2) + 6 = 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<em>x<\/em><sup>2<\/sup>\u00a0+\u00a0<em>kx<\/em> + 3 = 0<br \/>\n<\/strong>Comparing the given equation with\u00a0<em>ax<\/em><sup>2<\/sup>\u00a0+\u00a0<em>bx\u00a0<\/em>+\u00a0<em>c<\/em>\u00a0= 0, we get<br \/>\n<em>a<\/em>\u00a0= 2,\u00a0<em>b<\/em>\u00a0= k and\u00a0<em>c<\/em>\u00a0= 3<br \/>\nAs we know, discriminant =\u00a0<em>b<\/em><sup>2<\/sup>\u00a0\u2013 4<em>ac<br \/>\n<\/em>\u21d2 (<em>k<\/em>)<sup>2<\/sup>\u00a0\u2013 4(2) (3)<br \/>\n\u21d2 <em>k<\/em><sup>2<\/sup>\u00a0\u2013 24<br \/>\nFor equal roots, we know,<br \/>\nDiscriminant = 0<br \/>\n\u21d2 <em>k<\/em><sup>2<\/sup>\u00a0\u2013 24 = 0<br \/>\n\u21d2 <em>k<\/em><sup>2<\/sup>\u00a0= 24<br \/>\n\u21d2 k = \u00b1\u221a24 = \u00b12\u221a6 <\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii)\u00a0<em>kx<\/em>(<em>x<\/em>\u00a0\u2013 2) + 6 = 0<br \/>\n<\/strong><em>kx<\/em><sup>2<\/sup>\u00a0\u2013 2<em>kx<\/em>\u00a0+ 6 = 0<br \/>\nComparing the given equation with\u00a0<em>ax<\/em><sup>2<\/sup>\u00a0+\u00a0<em>bx\u00a0<\/em>+\u00a0<em>c<\/em>\u00a0= 0, we get<br \/>\n<em>a<\/em>\u00a0=\u00a0<em>k<\/em>,\u00a0<em>b<\/em>\u00a0= \u2013 2<em>k<\/em>\u00a0and\u00a0<em>c<\/em>\u00a0= 6<br \/>\nWe know, Discriminant =\u00a0<em>b<\/em><sup>2<\/sup>\u00a0\u2013 4<em>ac<br \/>\n<\/em>\u21d2 (\u2013 2<em>k<\/em>)<sup>2<\/sup>\u00a0\u2013 4 (<em>k<\/em>) (6)<br \/>\n\u21d2 4<em>k<sup>2<\/sup><\/em>\u00a0\u2013 24<em>k<br \/>\n<\/em>For equal roots, we know,<br \/>\n\u21d2 <em>b<\/em><sup>2<\/sup>\u00a0\u2013 4<em>ac<\/em>\u00a0= 0<br \/>\n\u21d2 4<em>k<\/em><sup>2<\/sup>\u00a0\u2013 24<em>k<\/em>\u00a0= 0<br \/>\n\u21d2 4<em>k<\/em>\u00a0(<em>k<\/em>\u00a0\u2013 6) = 0<br \/>\nEither 4<em>k<\/em>\u00a0= 0 or\u00a0<em>k<\/em>\u00a0= 6 = 0<br \/>\n\u21d2 <em>k<\/em>\u00a0= 0 or\u00a0<em>k<\/em>\u00a0= 6<br \/>\nHowever, if\u00a0<em>k<\/em>\u00a0= 0, then the equation will not have the terms \u2018<em>x<\/em><sup>2<\/sup>\u2018 and \u2018<em>x<\/em>\u2018.<br \/>\nTherefore, if this equation has two equal roots,\u00a0<em>k<\/em>\u00a0should be 6 only.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>3. Is it possible to design a rectangular mango grove whose length is twice its breadth and the area is 800 m<sup>2<\/sup>?\u00a0If so, find its length and breadth.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;\u00a0<\/strong>Let the breadth of the mango grove be\u00a0<em>l<\/em>.<br \/>\nThe length of the mango grove will be 2<em>l<\/em>.<br \/>\nArea of the mango grove = (2<em>l<\/em>) (<em>l<\/em>)= 2<em>l<\/em><sup>2<br \/>\n<\/sup>\u21d2 2<em>l<\/em><sup>2\u00a0<\/sup>= 800<br \/>\n\u21d2 <em>l<\/em><sup>2\u00a0<\/sup>= 800\/2<br \/>\n\u21d2 <em>l<\/em><sup>2\u00a0<\/sup>= 400<br \/>\n\u21d2 <em>l<\/em><sup>2\u00a0<\/sup>\u2013 400 =0<br \/>\nComparing the given equation with\u00a0<em>ax<\/em><sup>2<\/sup>\u00a0+\u00a0<em>bx\u00a0<\/em>+\u00a0<em>c<\/em>\u00a0= 0, we get<br \/>\n<em>a<\/em>\u00a0= 1,\u00a0<em>b<\/em>\u00a0= 0,\u00a0<em>c<\/em>\u00a0= 400<br \/>\nAs we know, discriminant =\u00a0<em>b<\/em><sup>2<\/sup>\u00a0\u2013 4<em>ac<br \/>\n<\/em>\u21d2 (0)<sup>2<\/sup>\u00a0\u2013 4 \u00d7 (1) \u00d7 ( \u2013 400) = 1600<br \/>\nHere,\u00a0<em>b<\/em><sup>2<\/sup>\u00a0\u2013 4<em>ac<\/em>\u00a0&gt; 0<br \/>\nThus, the equation will have real roots. And hence, the desired rectangular mango grove can be designed.<br \/>\n\u21d2 <em>l\u00a0<\/em>= \u00b120<br \/>\nAs we know, the value of length cannot be negative.<br \/>\nTherefore, the breadth of the mango grove = 20 m.<br \/>\nLength of the mango grove = 2 \u00d7 20 = 40 m.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>4. Is the following situation possible? If so, determine their present ages. The sum of the ages of the two friends is 20 years. Four years ago, the product of their age in years was 48.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211; <\/strong>Let\u2019s say the age of one friend is x years.<br \/>\nThen, the age of the other friend will be (20 \u2013 x) years.<br \/>\nFour years ago,<br \/>\nAge of first friend = (<em>x<\/em>\u00a0\u2013 4) years<br \/>\nAge of second friend = (20 \u2013\u00a0<em>x<\/em>\u00a0\u2013 4)\u00a0= (16 \u2013<em>\u00a0x<\/em>) years<br \/>\nAs per the given question, we can write,<br \/>\n\u21d2 (<em>x<\/em>\u00a0\u2013 4) (16 \u2013\u00a0<em>x<\/em>) = 48<br \/>\n\u21d2 16<em>x \u2013 x<\/em><sup>2<\/sup>\u00a0\u2013 64 + 4<em>x<\/em>\u00a0= 48<br \/>\n\u21d2 <em>\u2013 x<\/em><sup>2<\/sup>\u00a0+\u00a020<em>x \u2013\u00a0<\/em>112 = 0<br \/>\n\u21d2 <em>x<\/em><sup>2<\/sup>\u00a0\u2013\u00a020<em>x +\u00a0<\/em>112 = 0<br \/>\nComparing the equation with\u00a0<em>ax<\/em><sup>2<\/sup>\u00a0+\u00a0<em>bx\u00a0<\/em>+\u00a0<em>c<\/em> = 0, we get<br \/>\n<em>a<\/em>\u00a0=\u00a0<em>1<\/em>,\u00a0<em>b<\/em>\u00a0= -2<em>0<\/em>\u00a0and\u00a0<em>c<\/em> = 112<br \/>\nDiscriminant =\u00a0<em>b<\/em><sup>2<\/sup>\u00a0\u2013 4<em>ac<br \/>\n<\/em><em>= (-<\/em>20<em>)<\/em><sup>2<\/sup>\u00a0\u2013 4 \u00d7 112<br \/>\n= 400 \u2013 448 = -48<br \/>\n\u21d2 <em>b<\/em><sup>2<\/sup>\u00a0\u2013 4<em>ac\u00a0<\/em>&lt; 0<br \/>\nTherefore, there will be no real solution possible for the equations. Hence, the condition doesn\u2019t exist.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>5. Is it possible to design a rectangular park with a perimeter of 80 and an area of 400 m2? If so, find its length and breadth.<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solution &#8211;\u00a0 <\/strong>Let the length and breadth of the park be\u00a0<em>l\u00a0<\/em>and b.<br \/>\nPerimeter of the rectangular park = 2 (<em>l + b<\/em>) = 80<br \/>\n<em>So, l + b\u00a0<\/em>= 40<br \/>\nOr,\u00a0<em>b<\/em>\u00a0= 40 \u2013\u00a0<em>l<br \/>\n<\/em>Area of the rectangular park =\u00a0<em>l\u00d7b = l(40 \u2013 l) =\u00a0<\/em>40<em>l\u00a0<\/em>\u2013\u00a0<em>l<\/em><sup>2\u00a0<\/sup>= 400<br \/>\n<em>l<\/em><sup>2<\/sup><em>\u00a0<\/em>\u2013<sup>\u00a0\u00a0<\/sup>40<em>l<\/em>\u00a0+ 400<em>\u00a0<\/em>= 0, which is a quadratic equation.<br \/>\nComparing the equation with\u00a0<em>ax<\/em><sup>2<\/sup>\u00a0+\u00a0<em>bx\u00a0<\/em>+\u00a0<em>c<\/em>\u00a0= 0, we get<br \/>\n<em>a<\/em>\u00a0= 1,\u00a0<em>b<\/em>\u00a0= -40,\u00a0<em>c<\/em>\u00a0= 400<br \/>\nSince discriminant =\u00a0<em>b<\/em><sup>2<\/sup>\u00a0\u2013 4<em>ac<br \/>\n<\/em>\u21d2 <em>(-<\/em>40<em>)<\/em><sup>2<\/sup>\u00a0\u2013 4 \u00d7 400<br \/>\n\u21d2 1600 \u2013 1600 = 0<br \/>\nThus,<em>\u00a0b<\/em><sup>2<\/sup>\u00a0\u2013 4<em>ac\u00a0<\/em>= 0<br \/>\nTherefore, this equation has equal real roots. Hence, the situation is possible.<br \/>\nThe root of the equation,<br \/>\n<em>l<\/em>\u00a0= \u2013<em>b<\/em>\/2<em>a<br \/>\n<\/em><em>l<\/em>\u00a0= -(-40)\/2(1) = 40\/2 = 20<br \/>\nTherefore, the length of the rectangular park,\u00a0<em>l\u00a0<\/em>= 20 m<br \/>\nAnd the breadth of the park,\u00a0<em>b\u00a0<\/em>= 40 \u2013\u00a0<em>l\u00a0<\/em>= 40 \u2013 20 = 20 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 4.4 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":[1045,1069,1073,1044,1049,1048],"class_list":["post-6036","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-4-4-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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