Question

if alpha and beta are the zeros of the polynomial x^2-x-12=0 then form a quadratic equation whose zeros are 2alpha,2beta

Answer 1

$\alpha + \beta = - ( - 1) = 1 \\ \alpha \beta = - 12 \\ \\ let \: s \: and \: p \: denote \:the \: sum \: and \: product \: of \: required \: quadratic \: equation \\ \\ 2 \alpha \: and \: 2 \beta \: are the \: zeros \: of \: required \: equation \\ \\ s = 2 \alpha + 2 \beta = 2( \alpha + \beta ) = 2 \times 1 = 2 \\ \\ p = 2 \alpha \times 2 \beta = 4 \alpha \beta = 4 \times - 12 = - 48 \\ \\ required \: equation = \\ \\ x {}^{2} - ( 2 \alpha + 2 \beta ) + 4 \alpha \beta \\ = x {}^{2} - 2 - 48$

Swipe for the answer horizontally not visible.

Answer 2

x^2-x-12=0 => x^2-4x+3x-12=0 => x(x-4)+3(x-4) =0

=> (x-4)(x+3) =0 => x=4 or x=-3.

Let alpha=4 and beta=-3

Then, 2 alpha=(2*4)=8 and 2 beta=(-3*2)=-6.

Hence, the required equation

= x^2-(2 alpha+2 beta)x+2 alpha 2 beta

x^2- {8+(-6)}x + {8*(-6)} = x^2-2x-48.