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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.
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Solution
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.
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