Question

Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes  respectively:-  (i)  2/3 , -4 (ii) -1/root2, 1/root 2 (iii) root 5 , -2 (iv) -1/root 2, 2/3​

Answer 1

Answer:

$(i) \alpha + \beta = \frac{2}{3}, \: \alpha \beta = ( - 4) \\ Quadratic \: polynomial \\ = {x}^{2} - (\alpha + \beta )x + \alpha \beta \\ = {x}^{2} - \frac{2}{3} x + ( - 4) \\ = {x}^{2} - \frac{2x}{3} - 4 \\ multiply \: each \: term \: by \: 3 \\ = {x}^{2} \times 3 - \frac{2x}{3} \times 3 - 4 \times 3 \\ = 3 {x}^{2} - 2x - 12 \\ (ii) \alpha + \beta = \frac{( - 1)}{ \sqrt{2} } , \: \alpha \beta = \frac{1}{ \sqrt{2} } \\ Quadratic \: polynomial \\ = {x}^{2} - (\alpha + \beta )x + \alpha \beta \\ = {x}^{2} - \frac{( - 1)}{ \sqrt{2} } x + \frac{1}{ \sqrt{2} } \\ = {x}^{2} + \frac{x}{ \sqrt{2} } + \frac{1}{ \sqrt{2} } \\ multiply \: each \: term \: by \: \sqrt{2} \\ = {x}^{2} \times \sqrt{2} + \frac{x}{ \sqrt{2} } \times \sqrt{2} + \frac{1}{ \sqrt{2} } \times \sqrt{2} \\ = \sqrt{2} {x}^{2} + x + 1 \\ (iii) \alpha + \beta = \sqrt{5} , \: \alpha \beta = ( - 2) \\ Quadratic \: polynomial \\ = {x}^{2} - (\alpha + \beta )x + \alpha \beta \\ = {x}^{2} - \sqrt{5} x + ( - 2) \\ = {x}^{2} - \sqrt{5} x - 2 \\ (iv) \alpha + \beta = \frac{( - 1)}{ \sqrt{2} }, \: \alpha \beta = \frac{2}{3} \\ Quadratic \: polynomial \\ = {x}^{2} - (\alpha + \beta )x + \alpha \beta \\ = {x}^{2} - \frac{( - 1)}{ \sqrt{2} } x + \frac{2}{3} \\ = {x}^{2} + \frac{x}{ \sqrt{2} } + \frac{2}{3} \\ multiply \: each \: term \: by \: 3 \sqrt{2} \\ = {x}^{2} \times 3 \sqrt{2} + \frac{x}{ \sqrt{2} } \times 3 \sqrt{2} + \frac{2}{3} \times 3 \sqrt{2} \\ = 3 \sqrt{2} {x}^{2} + 3x + 2 \sqrt{2}$