Question

If sum of the square of zeroes of the polynomial $f\left(x\right)=x^2-8x+k$ is $40$, find the value of k.

• A. $14$
• B. $12$
• C. $6$
• D. $18$

Answer 1

Answer: (B)

$12$

We are given,

sum of the square of zeros of

polynomial $f\left(x\right)=x^2-8x+k$ is

40.

Let take $\alpha$ & $\beta$ are the roots of

$f\left(x\right)=x^2-8x+k$

here ${\alpha }^2+{\beta }^2=40$

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

Now, sum of zeros $\alpha +\beta =\frac{-b}{a}$

$\alpha +\beta =-(-8)=8$

& product of zeros $\alpha \beta =\frac{c}{a}$

$\alpha \beta =\frac{k}{1}=k$

so, $(\alpha +\beta {)}^2-2\alpha \beta =40$

$(8{)}^2-2(k)=40$

$64-40=2k$

$2k=24$

$k=12$