Question

If the zeroes of the polynomial $f\left(x\right)=x^3-12x^2+39x+k$ are in AP, the value of $k$ is

• A. $28$
• B. $-28$
• C. $30$
• D. $-30$

Answer 1

The correct option is B $-28$

We have,

$f\left(x\right)=x^3-12x^2+39x+k$

Since, roots of this equation are in $A.P.$

Let $a-d,a,a+d$ are roots.

Now, sum of roots = $\frac{-b}{a}$

$a-d+a+a+d=\frac{12}{1}$

$3a=12$

$a=4$

Sum of products of two consecutive roots = $\frac{c}{a}$

$(a-d)a+a(a+d)+(a-d)(a+d)=\frac{39}{1}$

$a^2-ad+a^2+ad+a^2-d^2=39$

$3a^2-d^2=39$

$3\times 16-d^2=39$

$48-d^2=39$

$d^2=$

$d=\pm 3$

Therefore, the roots are 1, 4, 7 or 7, 4, 1

Now, product of roots$=(a-d)a(a+d)=\frac{-d}{a}$

$1\times 4\times 7=-k$

$k=-28$

Hence, the value of $k$ is $-28$.

Hence, option b is the correct option.