Question

if sum of square of zeroes of quadratic polynomial x²-8x+k is 40, then find the value of k.​

Answer 1

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

$\sf \: Let \: \alpha \: and \: \beta \: be \: zeroes \: of \: f(x) = {x}^{2} - 8x + k$

We know that,

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

$\bf\implies \: \alpha + \beta = - \dfrac{( - 8)}{1} = 8$

And

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

$\bf\implies \: \alpha \beta = \dfrac{k}{1} = k$

Now,

According to statement,

It is given that,

$\rm :\longmapsto\: { \alpha }^{2} + { \beta }^{2} = 40$

$\rm :\longmapsto\: {( \alpha + \beta )}^{2} - 2 \alpha \beta = 40$

$\rm :\longmapsto\: {8}^{2} - 2k = 40$

$\rm :\longmapsto\:64 - 2k = 40$

$\rm :\longmapsto\:2k = 64 - 40$

$\rm :\longmapsto\:2k = 24$

$\bf\implies \:k \: = \: 12$

Additional Information :-

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

$\boxed{\bf\: \alpha - \beta = \sqrt{ {( \alpha }+\beta )^{2} - 4 \alpha \beta}}$