Question

If the sum of the cubes of zeroes of the polynomial
x3 - 5x2+k2x+8 is 71, then √k can be
(2) -42
(1) -√2
(3) 4/2
4'√2​

Answer 1

the answer is the d. part

Answer 2

Answer:4

Step-by-step explanation:

Given

$x^3-5x^2+k^2x+8=0$

sum of cubes of roots is 71

Let a, b and c be the roots of Equation

$a^3+b^3+c^3=71$

and we know

$a^3+b^3+c^3-3abc=\left ( a+b+c\right )\left [ \left ( a+b+c\right )^2-3\left ( ab+bc+ac\right )\right ]$

and from above equations

$a+b+c=5$

$abc=-8$

$ab+bc+ac=k^2$

Substitute

$71=3\cdot (-8)+5\left ( 5^2-3\times k^2\right )$

$15k^2=30$

$k=\sqrt{2}$