{"id":6011,"date":"2023-07-19T06:52:00","date_gmt":"2023-07-19T06:52:00","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=6011"},"modified":"2023-07-23T05:43:11","modified_gmt":"2023-07-23T05:43:11","slug":"ncert-solutions-class-10-maths-chapter-2-polynomials-ex-2-2","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-2-polynomials-ex-2-2\/","title":{"rendered":"NCERT Solutions Class 10 Maths Chapter 2 Polynomials Ex 2.2"},"content":{"rendered":"<h2 style=\"text-align: center;\"><strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">NCERT Solutions Class 10 Maths\u00a0<\/span><\/strong><br \/>\n<strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Chapter &#8211; 2 (Polynomials)\u00a0<\/span><\/strong><\/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; 2 Polynomials <\/strong>Exercise 2.2 has been provided here to help the students in solving the questions from this exercise.\u00a0<\/span><\/p>\n<p><span style=\"color: #ff0000;\"><strong>Chapter : 2 Polynomials<\/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-2-polynomials-ex-2-1\/\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 2.1<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-2-polynomials-ex-2-3\/\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 2.3<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-2-polynomials-ex-2-4\/\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 2.4<\/a><\/span><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 2.2<\/span><\/strong><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.<br \/>\n<\/strong><strong>(i) x<sup>2<\/sup> &#8211; 2x &#8211; 8\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(ii) 4s<sup>2<\/sup> &#8211; 4s + 1\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (iii) 6x<sup>2<\/sup>\u00a0&#8211; 3 &#8211; 7x<\/strong><\/span><br \/>\n<span style=\"color: #000000;\"><strong>(iv) 4u<sup>2<\/sup> + 8u\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(v) t<sup>2<\/sup> &#8211; 15\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(vi) 3x<sup>2<\/sup>\u00a0&#8211; x &#8211; 4<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solutions &#8211;<br \/>\n<\/strong><strong>(i) x<sup>2 <\/sup>\u2013 2x \u2013 8<br \/>\n<\/strong><strong>\u21d2<\/strong>x<sup>2 <\/sup>\u2013 4x + 2x \u2013 8<br \/>\n= x(x \u2013 4) + 2(x \u2013 4)<br \/>\n= (x &#8211; 4)(x + 2)<br \/>\nx = 4 , x = -2 are the zeroes of the polynomial.<br \/>\nThus, \u03b1 = 4, \u03b2 = -2<br \/>\nSum of zeroes = &#8211; coefficient of x \/ coefficient of x<sup>2<br \/>\n<\/sup>For\u00a0x<sup>2<\/sup>\u00a0&#8211; 2x &#8211; 8,<\/span><br \/>\n<span style=\"color: #000000;\">a = 1, b = &#8211; 2, c = &#8211; 8<\/span><br \/>\n<span style=\"color: #000000;\">\u03b1 + \u03b2 = &#8211; b \/ a<\/span><br \/>\n<span style=\"color: #000000;\">Here, <\/span><br \/>\n<span style=\"color: #000000;\">\u03b1 + \u03b2 = &#8211; 2 + 4 = 2 <\/span><br \/>\n<span style=\"color: #000000;\">&#8211; b \/ a = &#8211; (- 2)\/1 = 2<\/span><br \/>\n<span style=\"color: #000000;\">Hence, sum of the zeros \u03b1 + \u03b2 = &#8211; b\/a is verified.<\/span><br \/>\n<span style=\"color: #000000;\">Now, Product of zeroes = constant term \/ coefficient of x<sup>2<br \/>\n<\/sup>\u03b1 \u00d7 \u03b2 = c \/ a <\/span><br \/>\n<span style=\"color: #000000;\">Here, <\/span><br \/>\n<span style=\"color: #000000;\">\u03b1 \u00d7 \u03b2 = &#8211; 2 \u00d7 4 = &#8211; 8 <\/span><br \/>\n<span style=\"color: #000000;\">c \/ a = &#8211; 8 \/ 1 = &#8211; 8 <\/span><br \/>\n<span style=\"color: #000000;\">Hence, product of zeros \u03b1 \u00d7 \u03b2 = c \/ a is verified.<\/span><br \/>\n<span style=\"color: #000000;\">Thus,\u00a0x = 4,\u00a0-2 are the zeroes of the polynomial.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) 4s<sup>2 <\/sup>\u2013 4s + 1<br \/>\n<\/strong>\u21d2 4s<sup>2 <\/sup>\u2013 2s \u2013 2s + 1<br \/>\n= 2s(2s \u2013 1) \u2013 1(2s \u2013 1)<br \/>\n= (2s \u2013 1)(2s \u2013 1)<br \/>\ns = 1\/2, s = 1\/2 are the zeroes of the polynomial.<br \/>\nThus,\u00a0\u03b1 = 1\/2 and\u00a0\u03b2 = 1\/2<\/span><br \/>\n<span style=\"color: #000000;\">Now, let&#8217;s find the relationship between the zeroes and the coefficients.<\/span><br \/>\n<span style=\"color: #000000;\">For\u00a04s<sup>2<\/sup>\u00a0&#8211; 4s + 1,<\/span><br \/>\n<span style=\"color: #000000;\">a = 4, b = &#8211; 4 and c = 1<\/span><br \/>\n<span style=\"color: #000000;\">Sum of zeroes = &#8211; coefficient of s \/ coefficient of s<sup>2<br \/>\n<\/sup>\u03b1 + \u03b2 = &#8211; b\/a<\/span><br \/>\n<span style=\"color: #000000;\">Here,<\/span><br \/>\n<span style=\"color: #000000;\">\u03b1 + \u03b2 =\u00a01\/2 + 1\/2 = 1<\/span><br \/>\n<span style=\"color: #000000;\">&#8211; b \/ a = &#8211; (- 4) \/ 4 = 1<\/span><br \/>\n<span style=\"color: #000000;\">Hence, \u03b1 + \u03b2 = &#8211; b \/ a, verified<\/span><br \/>\n<span style=\"color: #000000;\">Now, Product of zeroes = constant term \/ coefficient of s<sup>2<br \/>\n<\/sup>\u03b1 \u00d7 \u03b2 = c \/ a<\/span><br \/>\n<span style=\"color: #000000;\">Here, <\/span><br \/>\n<span style=\"color: #000000;\">\u03b1 \u00d7 \u03b2 = 1\/2 \u00d7 1\/2 = 1\/4<\/span><br \/>\n<span style=\"color: #000000;\">c \/ a = 1\/4<\/span><br \/>\n<span style=\"color: #000000;\">Hence,\u00a0\u03b1 \u00d7 \u03b2 = c \/ a, verified.<\/span><br \/>\n<span style=\"color: #000000;\">Thus,\u00a0s = 1\/2, 1\/2 are the zeroes of the polynomial.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii) 6x<sup>2 <\/sup>\u2013 3 \u2013 7x<br \/>\n<\/strong>\u21d2 6x<sup>2 <\/sup>\u2013 7x \u2013 3<br \/>\n= 6x<sup>2 <\/sup>\u2013 9x + 2x \u2013 3<br \/>\n= 3x(2x \u2013 3) + 1(2x \u2013 3)<br \/>\n= (3x + 1)(2x \u2013 3)<br \/>\nx = 3\/2, x = &#8211; 1\/3 are the zeroes of the polynomial.<\/span><br \/>\n<span style=\"color: #000000;\">Thus,\u00a0\u03b1 = 3\/2 and\u00a0\u03b2 = -1\/3<\/span><br \/>\n<span style=\"color: #000000;\">Now, let&#8217;s find the relationship between the zeroes and the coefficients:<\/span><br \/>\n<span style=\"color: #000000;\">For\u00a06x<sup>2<\/sup>\u00a0&#8211; 3 &#8211; 7x,<\/span><br \/>\n<span style=\"color: #000000;\">a = 6, b = &#8211; 7 and c = &#8211; 3<\/span><br \/>\n<span style=\"color: #000000;\">Sum of zeroes = &#8211; coefficient of x \/ coefficient of x<sup>2<br \/>\n<\/sup>\u03b1 + \u03b2 = &#8211; b \/ a <\/span><br \/>\n<span style=\"color: #000000;\">Here, <\/span><br \/>\n<span style=\"color: #000000;\">\u03b1 + \u03b2 = 3\/2 + (-1\/3) = 7\/6 <\/span><br \/>\n<span style=\"color: #000000;\">&#8211; b \/ a = &#8211; (-7) \/ 6 = 7\/6 <\/span><br \/>\n<span style=\"color: #000000;\">Hence, \u03b1 + \u03b2 = &#8211; b\/a, verified<\/span><br \/>\n<span style=\"color: #000000;\">Now, Product of zeroes = constant term \/ coefficient of x<sup>2<br \/>\n<\/sup>\u03b1 \u00d7 \u03b2 = c \/ a<\/span><br \/>\n<span style=\"color: #000000;\">Here, <\/span><br \/>\n<span style=\"color: #000000;\">\u03b1 \u00d7 \u03b2 = 3\/2 \u00d7 (- 1\/3) = -1\/2 <\/span><br \/>\n<span style=\"color: #000000;\">c \/ a = (- 3) \/ 6 = -1\/2<\/span><br \/>\n<span style=\"color: #000000;\">Hence,\u00a0\u03b1 \u00d7 \u03b2 = c\/a, verified.<\/span><br \/>\n<span style=\"color: #000000;\">Thus, x = 3\/2, &#8211; 1\/3 are the zeroes of the polynomial.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iv) 4u<sup>2 <\/sup>+ 8u<br \/>\n<\/strong>\u21d2 4u(u + 2)<br \/>\nu = 0, u = &#8211; 2 are the zeroes of the polynomial<\/span><br \/>\n<span style=\"color: #000000;\">Thus,\u00a0\u03b1 = 0 and\u00a0\u03b2 = &#8211; 2<\/span><br \/>\n<span style=\"color: #000000;\">Now, let&#8217;s find the relationship between the zeroes and the coefficients<\/span><br \/>\n<span style=\"color: #000000;\">For\u00a04u<sup>2<\/sup>\u00a0+ 8u,<\/span><br \/>\n<span style=\"color: #000000;\">a = 4, b = 8, c = 0<\/span><br \/>\n<span style=\"color: #000000;\">Sum of zeroes = &#8211; coefficient of u \/ coefficient of u<sup>2<br \/>\n<\/sup>\u03b1 + \u03b2 = &#8211; b\/a<\/span><br \/>\n<span style=\"color: #000000;\">Here,<\/span><br \/>\n<span style=\"color: #000000;\">\u03b1 + \u03b2 =\u00a00 + (- 2) = &#8211; 2<\/span><br \/>\n<span style=\"color: #000000;\">&#8211; b \/ a = &#8211; (8) \/ 4 =\u00a0&#8211; 2<\/span><br \/>\n<span style=\"color: #000000;\">Hence, \u03b1 + \u03b2 = &#8211; b \/ a, verified<\/span><br \/>\n<span style=\"color: #000000;\">Now, Product of zeroes = constant term \/ coefficient of u<sup>2<br \/>\n<\/sup>\u03b1 \u00d7 \u03b2 = c\/a<\/span><br \/>\n<span style=\"color: #000000;\">Here,<\/span><br \/>\n<span style=\"color: #000000;\">\u03b1 \u00d7 \u03b2 = 0 \u00d7 (- 2) = 0<\/span><br \/>\n<span style=\"color: #000000;\">c \/ a = 0 \/ 4 = 0<\/span><br \/>\n<span style=\"color: #000000;\">Hence,\u00a0\u03b1 \u00d7 \u03b2 = c \/ a, verified.<\/span><br \/>\n<span style=\"color: #000000;\">Thus, u = 0, &#8211; 2\u00a0are the zeroes of the polynomial.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(v) t<sup>2 <\/sup>\u2013 15<br \/>\n<\/strong>\u21d2 t<sup>2<\/sup> = 15<br \/>\n\u21d2 t = \u00b1\u221a15<br \/>\nt = -\u221a15, t = \u221a15 are the zeroes of the polynomial.<\/span><br \/>\n<span style=\"color: #000000;\">Thus,\u00a0\u03b1 =\u00a0-\u221a15 and\u00a0\u03b2 =\u00a0\u221a15<\/span><br \/>\n<span style=\"color: #000000;\">Now, let&#8217;s find the relationship between the zeroes and the coefficients<\/span><br \/>\n<span style=\"color: #000000;\">For\u00a0t<sup>2<\/sup>\u00a0&#8211; 15,<\/span><br \/>\n<span style=\"color: #000000;\">a = 1, b = 0, c = -15<\/span><br \/>\n<span style=\"color: #000000;\">Sum of zeroes = &#8211; coefficient of t \/ coefficient of t<sup>2<br \/>\n<\/sup>\u03b1 + \u03b2 = &#8211; b \/ a<\/span><br \/>\n<span style=\"color: #000000;\">Here,<\/span><br \/>\n<span style=\"color: #000000;\">\u03b1 + \u03b2 = -\u221a15 + \u221a15 = 0<\/span><br \/>\n<span style=\"color: #000000;\">&#8211; b \/ a = &#8211; 0 \/ 1 = 0<\/span><br \/>\n<span style=\"color: #000000;\">Hence, \u03b1 + \u03b2 = &#8211; b \/ a, verified<\/span><br \/>\n<span style=\"color: #000000;\">Now, Product of zeroes = constant term \/ coefficient of t<sup>2<br \/>\n<\/sup>\u03b1 \u00d7 \u03b2 = c \/ a<\/span><br \/>\n<span style=\"color: #000000;\">Here,<\/span><br \/>\n<span style=\"color: #000000;\">\u03b1 \u00d7 \u03b2 = -\u221a15 \u00d7 \u221a15 = -15<\/span><br \/>\n<span style=\"color: #000000;\">c \/ a = -15 \/ 1 = -15<\/span><br \/>\n<span style=\"color: #000000;\">Hence, \u03b1 \u00d7 \u03b2 = c \/ a, verified.<\/span><br \/>\n<span style=\"color: #000000;\">Thus, t\u00a0= -\u221a15, \u221a15 are the zeroes of the polynomial.<\/span><\/p>\n<p><span style=\"color: #000000;\"><strong>(vi)\u00a03x<sup>2<\/sup>\u00a0&#8211; x &#8211; 4<br \/>\n<\/strong>\u21d2 3x<sup>2<\/sup>\u00a0&#8211; x &#8211; 4 = 0<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 3x<sup>2<\/sup>\u00a0&#8211; 4x + 3x &#8211; 4 = 0<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 x (3x &#8211; 4) + 1(3x &#8211; 4) = 0<\/span><br \/>\n<span style=\"color: #000000;\">\u21d2 (x + 1)(3x &#8211; 4) = 0<\/span><br \/>\n<span style=\"color: #000000;\">x = &#8211; 1, x = 4\/3 are the zeroes of the polynomial.<\/span><br \/>\n<span style=\"color: #000000;\">Thus,\u00a0\u03b1 = &#8211; 1 and\u00a0\u03b2 = 4\/3<\/span><br \/>\n<span style=\"color: #000000;\">Now, let&#8217;s find the relationship between the zeroes and the coefficients<\/span><br \/>\n<span style=\"color: #000000;\">For\u00a03x<sup>2<\/sup>\u00a0&#8211; x &#8211; 4,<\/span><br \/>\n<span style=\"color: #000000;\">a = 3, b = &#8211; 1, c = &#8211; 4<\/span><br \/>\n<span style=\"color: #000000;\">Sum of zeroes = &#8211; coefficient of x \/ coefficient of x<sup>2<br \/>\n<\/sup>\u03b1 + \u03b2 = &#8211; b \/ a<\/span><br \/>\n<span style=\"color: #000000;\">Here,<\/span><br \/>\n<span style=\"color: #000000;\">\u03b1 + \u03b2 = &#8211; 1 + 4\/3\u00a0 = 1\/3<\/span><br \/>\n<span style=\"color: #000000;\">&#8211; b \/ a = &#8211; (-1) \/ 3 = 1\/3<\/span><br \/>\n<span style=\"color: #000000;\">Hence, \u03b1 + \u03b2 = &#8211; b \/ a, verified<\/span><br \/>\n<span style=\"color: #000000;\">Now, Product of zeroes = constant term \/ coefficient of x<sup>2<br \/>\n<\/sup>\u03b1 \u00d7 \u03b2 = c\/a<\/span><br \/>\n<span style=\"color: #000000;\">Here,<\/span><br \/>\n<span style=\"color: #000000;\">\u03b1 \u00d7 \u03b2 = &#8211; 1 \u00d7 (4\/3) = &#8211; 4\/3<\/span><br \/>\n<span style=\"color: #000000;\">c \/ a = &#8211; 4\/3<\/span><br \/>\n<span style=\"color: #000000;\">Hence, \u03b1 \u00d7 \u03b2 = c \/ a, verified.<\/span><br \/>\n<span style=\"color: #000000;\">Thus, x\u00a0= &#8211; 1,\u00a04\/3 are the zeroes of the polynomial.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes, respectively.<br \/>\n<\/strong><strong>(i) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{1}{4}}\" alt=\"\\mathbf{\\frac{1}{4}}\" align=\"absmiddle\" \/>, &#8211; 1\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (ii) \u221a2, <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{1}{3}}\" alt=\"\\mathbf{\\frac{1}{3}}\" align=\"absmiddle\" \/>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (iii) 0, \u221a5<\/strong><\/span><\/p>\n<p>(iv) 1, 1\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (v) &#8211; <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{1}{4}}\" alt=\"\\mathbf{\\frac{1}{4}}\" align=\"absmiddle\" \/>, <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{1}{4}}\" alt=\"\\mathbf{\\frac{1}{4}}\" align=\"absmiddle\" \/>\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0<strong>(vi) 4, 1<\/strong><\/p>\n<p><span style=\"color: #000000;\"><strong>Solution &#8211;\u00a0<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(i) <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{1}{4}}\" alt=\"\\mathbf{\\frac{1}{4}}\" align=\"absmiddle\" \/>, &#8211; 1<br \/>\n<\/strong>Sum of zeroes = \u03b1+\u03b2 = 1\/4<br \/>\nProduct of zeroes = \u03b1 \u03b2 = -1<\/span><br \/>\n<span style=\"color: #000000;\">We know that the general equation of a quadratic polynomial is:<\/span><br \/>\n<span style=\"color: #000000;\">x<sup>2<\/sup>\u00a0&#8211; (sum of roots) x + (product of roots)<\/span><br \/>\n<span style=\"color: #000000;\">x<sup>2<\/sup>\u2013(\u03b1 + \u03b2)x + \u03b1\u03b2 = 0<\/span><br \/>\n<span style=\"color: #000000;\">x<sup>2<\/sup>\u00a0&#8211; (1\/4)x + (- 1)<\/span><br \/>\n<span style=\"color: #000000;\">x<sup>2<\/sup>\u00a0&#8211; (1\/4)x &#8211; 1<\/span><br \/>\n<span style=\"color: #000000;\">4x<sup>2<\/sup>\u2013x-4 = 0<br \/>\nThus<strong>, <\/strong>4x<sup>2<\/sup>\u2013x\u20134 is the quadratic polynomial.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) \u221a2, <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{1}{3}}\" alt=\"\\mathbf{\\frac{1}{3}}\" align=\"absmiddle\" \/><br \/>\n<\/strong>Sum of zeroes = \u03b1 + \u03b2 =\u221a2<br \/>\nProduct of zeroes = \u03b1 \u03b2 = 1\/3<br \/>\nWe know that the general equation of a quadratic polynomial is:<br \/>\nx<sup>2<\/sup>\u00a0&#8211; (sum of roots) x + (product of roots)<br \/>\nx<sup>2 <\/sup>\u2013 (\u03b1 + \u03b2)x + \u03b1\u03b2 = 0<br \/>\nx<sup>2<\/sup>\u00a0\u2013(\u221a2)x + (1\/3) = 0<br \/>\n3x<sup>2 <\/sup>&#8211; 3\u221a2x + 1 = 0<br \/>\nThus, 3x<sup>2 <\/sup>&#8211; 3\u221a2x + 1 is the quadratic polynomial.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii) 0, \u221a5<br \/>\n<\/strong>Sum of zeroes = \u03b1 + \u03b2 = 0<br \/>\nProduct of zeroes = \u03b1 \u03b2 = \u221a5<br \/>\nWe know that the general equation of a quadratic polynomial is:<br \/>\nx<sup>2<\/sup>\u00a0&#8211; (sum of roots) x + (product of roots)<br \/>\nx<sup>2 <\/sup>\u2013 (\u03b1 + \u03b2)x + \u03b1\u03b2 = 0<br \/>\nx<sup>2 <\/sup>\u2013 (0)x + \u221a5= 0<br \/>\nThus, x<sup>2<\/sup>+\u221a5 is the quadratic polynomial.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iv) 1, 1<br \/>\n<\/strong>Sum of zeroes = \u03b1+\u03b2 = 1<br \/>\nProduct of zeroes = \u03b1 \u03b2 = 1<br \/>\nWe know that the general equation of a quadratic polynomial is:<br \/>\nx<sup>2<\/sup>\u00a0&#8211; (sum of roots) x + (product of roots)<br \/>\nx<sup>2 <\/sup>\u2013 (\u03b1 + \u03b2)x + \u03b1\u03b2 = 0<br \/>\nx<sup>2 <\/sup>\u2013 x + 1 = 0<br \/>\nThus, x<sup>2 <\/sup>\u2013 x + 1 is the quadratic polynomial.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(v) &#8211; <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{1}{4}}\" alt=\"\\mathbf{\\frac{1}{4}}\" align=\"absmiddle\" \/>, <img decoding=\"async\" src=\"https:\/\/latex.codecogs.com\/gif.latex?\\mathbf{\\frac{1}{4}}\" alt=\"\\mathbf{\\frac{1}{4}}\" align=\"absmiddle\" \/><br \/>\n<\/strong>Sum of zeroes = \u03b1+\u03b2 = -1\/4<br \/>\nProduct of zeroes = \u03b1 \u03b2 = 1\/4<br \/>\nWe know that the general equation of a quadratic polynomial is:<br \/>\nx<sup>2<\/sup>\u00a0&#8211; (sum of roots) x + (product of roots)<br \/>\nx<sup>2 <\/sup>\u2013 (\u03b1 + \u03b2)x + \u03b1\u03b2 = 0<br \/>\nx<sup>2 <\/sup>\u2013 (-1\/4)x + (1\/4) = 0<br \/>\n4x<sup>2<\/sup>+x+1 = 0<br \/>\nThus, 4x<sup>2 <\/sup>+ x + 1 is the quadratic polynomial.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(vi) 4, 1<br \/>\n<\/strong>Sum of zeroes = \u03b1+\u03b2 =4<br \/>\nProduct of zeroes = \u03b1\u03b2 = 1<br \/>\nWe know that the general equation of a quadratic polynomial is:<br \/>\nx<sup>2<\/sup>\u00a0&#8211; (sum of roots) x + (product of roots)<br \/>\nx<sup>2 <\/sup>\u2013 (\u03b1 + \u03b2)x + \u03b1\u03b2 = 0<br \/>\nx<sup>2 <\/sup>\u2013 4x + 1 = 0<br \/>\nThus, x<sup>2 <\/sup>\u2013 4x + 1 is the quadratic polynomial.<\/span><\/p>\n<p><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><\/p>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 10 Maths\u00a0 Chapter &#8211; 2 (Polynomials)\u00a0 The NCERT Solutions in English Language for Class 10 Mathematics Chapter &#8211; 2 Polynomials Exercise 2.2 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter : 2 Polynomials NCERT Class 10 Maths Solution Ex &#8211; 2.1 NCERT Class 10 [&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":[1045,1053,1055,1044,1049,1048],"class_list":["post-6011","post","type-post","status-publish","format-standard","hentry","category-class-10-maths","tag-class-10-ncert-mathematics-solutions","tag-ncert-class-10-mathematics-chapter-2-polynomials-solutions","tag-ncert-class-10-mathematics-exercise-2-2-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) - https:\/\/yoast.com\/product\/yoast-seo-premium-wordpress\/ -->\n<title>NCERT 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