Question

20.A and B are zeroes of x2-
6x+k.what is k if 3A+2B=20​

Answer 1

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

Given that

$\rm :\longmapsto\:A \: and \: B \: are \: zeroes \: of \: {x}^{2} - 6x + k$

We know,

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

$\bf\implies \:A + B = - \dfrac{( - 6)}{1} = 6 - - - (1)$

And

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

$\bf\implies \:AB = \dfrac{k}{1} = k - - - (2)$

Also,

Given that

$\rm :\longmapsto\:3A + 2B = 20$

$\rm :\longmapsto\:A + 2A + 2B = 20$

$\rm :\longmapsto\:A + 2(A + B) = 20$

$\rm :\longmapsto\:A + 2(6) = 20 \: \: \: \: \: \: \: \: \: \: \{ \because \:A + B = 6 \}$

$\rm :\longmapsto\:A + 12 = 20$

$\rm :\longmapsto\:A = 20 - 12$

$\bf\implies \:A = 8 - - - (3)$

On substituting A = 8 in equation (1), we get

$\rm :\longmapsto\:8 + B = 6$

$\rm :\longmapsto \: B = 6 - 8$

$\bf\implies \:B = - 2 - - - (4)$

On substituting the values of A and B in equation (2), we get

$\rm :\longmapsto\:k = 8 \times ( - 2)$

$\bf\implies \:k \: = \: - \: 16$

Additional information :-

$\rm :\longmapsto\: {A}^{2} + {B}^{2} = {(A + B)}^{2} - 2AB$

$\rm :\longmapsto\: {A}^{3} + {B}^{3} = {(A + B)}^{3} - 3AB(A + B)$

$\rm :\longmapsto\: {(A - B)}^{2} = {(A + B)}^{2} - 4AB$