Question

find the quadratic polynomials whose sum and products of the zeroes are -2√3 and -9. Also find its zeroes.​

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that

$\rm :\longmapsto\: \alpha \: and \: \beta \: be \: the \: zeroes \: of \: polynomial \: f(x).$

It is given that

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

and

$\rm :\longmapsto\: \alpha \beta = - \: 9$

So,

The required Quadratic polynomial is

$\rm :\longmapsto\:f(x) = {x}^{2} - ( \alpha + \beta)x + \alpha \beta$

On substitute the values from above,

$\rm :\longmapsto\:f(x) = {x}^{2} + 2 \sqrt{3}x - 9$

$\rm :\longmapsto\:f(x) = {x}^{2} + 3 \sqrt{3}x - \sqrt{3}x - 9$

can be again rewrite as

$\rm :\longmapsto\:f(x) = {x}^{2} + 3 \sqrt{3}x - \sqrt{3}x - 3 \times 3$

can be again rewrite as

$\rm :\longmapsto\:f(x) = {x}^{2} + 3 \sqrt{3}x - \sqrt{3}x - 3 \times \sqrt{3} \times \sqrt{3}$

$\rm :\longmapsto\:f(x) = x(x + 3 \sqrt{3} ) - \sqrt{3}(x + 3 \sqrt{3})$

$\rm :\longmapsto\:f(x) = (x + 3 \sqrt{3})(x - \sqrt{3})$

So, the zeroes of f(x) are

$\rm :\longmapsto\:x = - 3 \sqrt{3} \: \: \: or \: \: \sqrt{3}$

More to know

$\rm :\longmapsto\: \alpha \: and \beta \: are \: zeroes \: of \: a {x}^{2} + bx + c, \: then \:$

$\boxed{{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

and

$\boxed{{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$

Answer 2

Answer:

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