Question

If α and β are the zeroes of quadratic polynomial such that 3α -2β = 8 and 2α + 3β = 14, then form the family of polynomials.

Answer 1

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

Given that,

$\rm :\longmapsto\:3 \alpha - 2\beta = 8 - - - (1)$

and

$\rm :\longmapsto\:2 \alpha + 3\beta = 14 - - - (2)$

Multiply equation (1) by 3 and equation (2) by 2, we get

$\rm :\longmapsto\:9 \alpha - 6\beta = 24 - - - (3)$

$\rm :\longmapsto\:4 \alpha + 6\beta = 28 - - - (4)$

On adding equation (3) and (4), we get

$\rm :\longmapsto\:13 \alpha = 52$

$\bf\implies \: \alpha = 4 - - - (5)$

On substituting the value in equation (2), we get

$\rm :\longmapsto\:2 \times 4 + 3 \beta = 14$

$\rm :\longmapsto\:8 + 3 \beta = 14$

$\rm :\longmapsto\: 3 \beta = 14 - 8$

$\rm :\longmapsto\: 3 \beta = 6$

$\bf\implies \: \beta = 2 - - - (6)$

Now,

Consider,

$\rm :\longmapsto\: \alpha + \beta = 4 + 2 = 6$

and

$\rm :\longmapsto\: \alpha\beta = 4 \times 2 = 8$

So, required polynomial is given by

$\rm :\longmapsto\:f(x) = k\bigg( {x}^{2} - ( \alpha + \beta)x + \alpha\beta \bigg) \: where \: k \ne \: 0$

$\bf :\longmapsto\:f(x) = k\bigg( {x}^{2} - 6x + 8\bigg) \: where \: k \ne \: 0$

Additional Information :-

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

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

$\green{ \boxed{ \bf \: { \alpha }^{2} + { \beta }^{2} = {( \alpha + \beta) }^{2} - 2 \alpha \beta\: }}$

$\green{ \boxed{ \bf \: { \alpha }^{3} + { \beta }^{3} = {( \alpha + \beta) }^{3} - 3\alpha \beta( \alpha+ \beta) \: }}$