Question

find a quadratic polynomial whose sum and product respectively of the zeroes are -2√3,-9 also find the zeroes of theses polynomial
by factorisation​

Answer 1

Answer:

$\alpha + \beta = - 2 \sqrt{3} \\ \alpha \beta = - 9 \\ {( \alpha - \beta )}^{2} = {( \alpha + \beta )}^{2} - 4 \alpha \beta \\ {( \alpha - \beta )}^{2} = {( - 2 \sqrt{3} )}^{2} - 4( - 9) \\ = 12 + 36 \\ {( \alpha - \beta )}^{2} = 48 \\ \alpha - \beta = \sqrt{48} = 4 \sqrt{3} \\ \alpha + \beta + \alpha - \beta = - 2 \sqrt{3} + 4 \sqrt{3 } \\ 2 \alpha = 2 \sqrt{3} \\ \alpha = \sqrt{3} \\ \sqrt{3} + \beta = - 2 \sqrt{3} \\ \beta = - 2 \sqrt{3} - \sqrt{3} \\ \beta = - 3 \sqrt{ 3} \\ the \: roots \: are \: \sqrt{3} and \: - 3 \sqrt{3}$

Answer 2

$\alpha + \beta = - 2 \sqrt{3}$

$\alpha \beta = - 9$

wkt, general form of an equation in the terms of sum and product of the roots is

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

therefore, the equation will be

x²-(-2√3)+(-9)

= x²+2√3x-9

x²+2√3x-9=0

x²-√3x+3√3x-9=0

x(x-√3)+3√3(x-√3)

(x+3√3)(x-√3)

x= -3√3, +√3

The equation of the quadratic polynomial whose sum and product of roots are -2√3 and -9 respectively is

x²+2√3x-9 and its roots are -3√3 and +√3