{"id":10511,"date":"2022-04-23T19:57:59","date_gmt":"2022-04-23T14:27:59","guid":{"rendered":"https:\/\/mathemerize.com\/?p=10511"},"modified":"2022-04-23T19:58:56","modified_gmt":"2022-04-23T14:28:56","slug":"find-a-quadratic-polynomial-whose-sum-of-zeroes-and-product-of-zeroes-are-respectively","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/find-a-quadratic-polynomial-whose-sum-of-zeroes-and-product-of-zeroes-are-respectively\/","title":{"rendered":"Find a quadratic polynomial whose sum of zeroes and product of zeroes are respectively."},"content":{"rendered":"
Question<\/strong> : Find a quadratic polynomial whose sum of zeroes and product of zeroes are respectively.<\/p>\n (i)<\/strong>\u00a0 \\(1\\over 4\\), -1<\/p>\n (ii)<\/strong>\u00a0 \\(\\sqrt{2}\\), \\(1\\over 3\\)<\/p>\n (iii)<\/strong>\u00a0 0, \\(\\sqrt{5}\\)<\/p>\n (iv)<\/strong>\u00a0 1, 1<\/p>\n (v)\u00a0<\/strong> \\(-1\\over 4\\), \\(1\\over 4\\)<\/p>\n (vi)<\/strong>\u00a0 4, 1<\/p>\n Solution<\/strong> : Let the polynomial be \\(ax^2 + bx + c\\) and its zeroes be \\(\\alpha\\) and \\(\\beta\\).<\/p>\n (i)<\/strong>\u00a0 Here, \\(\\alpha\\) + \\(\\beta\\) = \\(1\\over 4\\)\u00a0 and \\(\\alpha\\).\\(\\beta\\) = -1<\/p>\n Thus, the polynomial formed = \\(x^2\\) – (sum of zeroes) x + Product of zeroes<\/p>\n = \\(x^2\\) – (\\(1\\over 4\\))x – 1<\/p>\n = \\(4x^2 – x – 4\\)<\/p>\n (ii)<\/strong>\u00a0 Here, \\(\\alpha\\) + \\(\\beta\\) = \\(\\sqrt{2}\\) \u00a0and \\(\\alpha\\).\\(\\beta\\) = \\(1\\over 3\\)<\/p>\n Thus, the polynomial formed = \\(x^2\\) – (sum of zeroes) x + Product of zeroes<\/p>\n = \\(x^2\\) – (\\(\\sqrt{2}\\))x + \\(1\\over 3\\)<\/p>\n = \\(3x^2 – 3\\sqrt{2}x + 1\\)\u00a0<\/p>\n (iii)<\/strong>\u00a0 Here, \\(\\alpha\\) + \\(\\beta\\) = 0\u00a0 and \\(\\alpha\\).\\(\\beta\\) = \\(\\sqrt{5}\\)\u00a0<\/p>\n Thus, the polynomial formed = \\(x^2\\) – (sum of zeroes) x + Product of zeroes<\/p>\n = \\(x^2\\) – (0)x + \\(\\sqrt{5}\\)\u00a0<\/p>\n = \\(x^2 + \\sqrt{5}\\)<\/p>\n (iv)<\/strong>\u00a0 Here, \\(\\alpha\\) + \\(\\beta\\) = 1\u00a0 and \\(\\alpha\\).\\(\\beta\\) = 1<\/p>\n Thus, the polynomial formed = \\(x^2\\) – (sum of zeroes) x + Product of zeroes<\/p>\n = \\(x^2\\) – (1)x + 1<\/p>\n = \\(x^2 – x + 1\\)<\/p>\n (v)<\/strong>\u00a0 Here, \\(\\alpha\\) + \\(\\beta\\) = \\(-1\\over 4\\)\u00a0 and \\(\\alpha\\).\\(\\beta\\) = \\(1\\over 4\\)\u00a0<\/p>\n Thus, the polynomial formed = \\(x^2\\) – (sum of zeroes) x + Product of zeroes<\/p>\n = \\(x^2\\) – (\\(-1\\over 4\\))x + \\(1\\over 4\\)\u00a0<\/p>\n = \\(4x^2 + x + 1\\)\u00a0<\/p>\n (vi)<\/strong>\u00a0 Here, \\(\\alpha\\) + \\(\\beta\\) = 4\u00a0 and \\(\\alpha\\).\\(\\beta\\) = 1<\/p>\n Thus, the polynomial formed = \\(x^2\\) – (sum of zeroes) x + Product of zeroes<\/p>\n = \\(x^2\\) – (4)x + 1<\/p>\n = \\(x^2 – 4x + 1\\)<\/p>\n\n\n <\/p>\n","protected":false},"excerpt":{"rendered":" Question : Find a quadratic polynomial whose sum of zeroes and product of zeroes are respectively. (i)\u00a0 \\(1\\over 4\\), -1 (ii)\u00a0 \\(\\sqrt{2}\\), \\(1\\over 3\\) (iii)\u00a0 0, \\(\\sqrt{5}\\) (iv)\u00a0 1, 1 (v)\u00a0 \\(-1\\over 4\\), \\(1\\over 4\\) (vi)\u00a0 4, 1 Solution : Let the polynomial be \\(ax^2 + bx + c\\) and its zeroes be \\(\\alpha\\) and …<\/p>\n