Question

Find the values of k for which the quadratic equation (k + 4) x² + (k + 1) x + 1 = 0 has equal roots. Also find these roots.

Answer 1

Given quadratic equation:
(k+4)x²+(k+1)x+1=0.

Since the given quadratic equation has equal roots, its discriminant should be zero.

∴ D = 0
⇒ (k+1)²−4 × (k+4) × 1=0
⇒k²+2k+1−4k−16=0
⇒k²−2k−15=0
⇒k²−5k+3k−15=0
⇒(k−5)(k+3)=0
⇒k−5=0 or k+3=0
⇒k=5 or −3
Thus, the values of k are 5 and −3.

For k = 5:
(k+4)x²+(k+1) x+1=0
⇒9x²+6x+1=0
⇒(3x)²+2(3x)+1=0
⇒(3x+1)²=0
⇒x=−1/3, −1/3
For k = −3:
(k+4)x²+(k+1)x+1=0
⇒x²−2x+1=0
⇒(x−1)²=0
⇒x=1,1

Thus, the equal root of the given quadratic equation is either 1 or −13.