Question

Find a quadratic polynomial each with the given numbers as the sum and product of its
zeroes respectively.
(ii) √12,1/3​

Answer 1

EXPLANATION.

$\sf \: sum \: of \: zeroes \: of \: quadratic \: equation \: = \sqrt{12}$

$\sf \: products \: of \: zeroes \: of \: quadratic \: equation \: = \dfrac{1}{3}$

As we know that,

Formula of quadratic equation.

$\sf \: {x}^{2} - ( \alpha + \beta )x + \alpha \beta$

Sum of zeroes of quadratic equation.

$\sf \: \alpha + \beta = \dfrac{ - b}{a}$

$\sf \: \alpha + \beta = \sqrt{12}$

Products of zeroes of quadratic equation.

$\sf \: \alpha \beta = \dfrac{c}{a}$

$\sf \: \alpha \beta = \dfrac{1}{3}$

Put the value in equation, we get.

$\sf \: {x}^{2} - ( \sqrt{12} )x + \dfrac{1}{3} = 0$

$\sf \: 3 {x}^{2} - 3 \sqrt{12} x + 1 = 0$

Answer 2

EXPLANATION.

$\sf \: sum \: of \: zeroes \: of \: quadratic \: equation \: = \sqrt{12}$

$\sf \: products \: of \: zeroes \: of \: quadratic \: equation \: = \dfrac{1}{3}$

As we know that,

Formula of quadratic equation.

$\sf \: {x}^{2} - ( \alpha + \beta )x + \alpha \beta$

Sum of zeroes of quadratic equation.

$\sf \: \alpha + \beta = \dfrac{ - b}{a}$

$\sf \: \alpha + \beta = \sqrt{12}$

Products of zeroes of quadratic equation.

$\sf \: \alpha \beta = \dfrac{c}{a}$

$\sf \: \alpha \beta = \dfrac{1}{3}$

Put the value in equation, we get.

$\sf \: {x}^{2} - ( \sqrt{12} )x + \dfrac{1}{3} = 0$

$\sf \: 3 {x}^{2} - 3 \sqrt{12} x + 1 = 0$