Question

If alpha and beta are the zeros of polynomial p(x) - x² - 2x - 8 ,then form a quadratic polynomial whose zeroes are 3(alpha) & 3(beta)

Answer 1

SOLUTION:-

Given,

${x}^{2} - 2x - 8 = 0$

And zeroes are:

$3 \alpha \: and \: 3 \beta$

So,

$\alpha + \beta = \frac{ - b}{a} = \frac{ - ( - 2)}{1} \\ \\ = > 2 \\ \\ \alpha \beta = \frac{c}{a} = \frac{ - 8}{1} \\ \\ = > - 8 \\ \\ = > 3 \alpha + 3 \beta = 3( \alpha + \beta ) \\ \\ = > 3(2) = 6 \\ \\ = > 3 \alpha \times 3 \beta = 9 \alpha \beta \\ \\ = > 9( - 8) = - 72$

Required quadratic polynomial:

${x}^{2} - (3\alpha + 3 \beta )x + (3 \alpha ).(3 \beta ) = 0 \\ \\ = > {x}^{2} - 6x - 72 = 0$

Hope it helps ☺️

Answer 2

α,β are the zeros of x²-2x-8=0. then, α+β=-(-2/1)=2 and

α×β=-8/1=-8

now, 3α+3β=3(α+β)=3×2=6 and 3α×3β=9αβ=9×-8=-72

the required quadratic polynomial is:

x²-(sum of the roots)x+product of the roots=0

or, x²-6x-72=0