For what values of K will the quadratic (4x+1)x2+(k+1)x+1=0 have equal roots.
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For what values of K will the quadratic (4x+1)x2+(k+1)x+1=0 have equal roots.
For what values of K will the quadratic (4x+1)x2+(k+1)x+1=0 have equal roots.
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.