Two roots of the cubic x3+3 x2+k x-12=0 are real and unequal but have the same absolute value. The value of k is

Question

Two roots of the cubic x3+3 x2+k x-12=0 are real and unequal but have the same absolute value. The value of k is (A) 4 (B) -4 (C) 6 (D) -9

Tardigrade Admin · Accepted Answer

Let the roots are r,-r and t Hence, r - r + t =-3 ⇒ t =-3. Now, product of the roots is (r)(-r)(t)=12 -r2(-3)=12 ⇒ r2=4 Hence, roots are 2,-2 and -3 Now, sum of the product taken two at a time is r(-r)+(-r) t+t r=k -4-6+6=k ⇒ k=-4.

Q. Two roots of the cubic x3+3x2+kx−12=0 are real and unequal but have the same absolute value. The value of k is

4

-4

6

-9

Let the roots are r,−r and t Hence, r−r+t=−3⇒t=−3. Now, product of the roots is (r)(−r)(t)=12 −r2(−3)=12⇒r2=4 Hence, roots are 2,−2 and -3 Now, sum of the product taken two at a time is r(−r)+(−r)t+tr=k −4−6+6=k⇒k=−4.

Q. Two roots of the cubic $x^{3} + 3 x^{2} + k x - 12 = 0$ are real and unequal but have the same absolute value. The value of $k$ is