Find the sum of values of a for which the equation (3a + 1)x^2+ 2(a + 1)x + 1 = 0 has real and equal roots.
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Answer: Given equation is y=(3k+1)x
2
+2(k+1)x+1=0
Also, it is given that the equation has equal roots.
Then D=0⇒b
2
−4ac=0
a=3k+1 b=2(k+1) c=1
b
2
−4ac=[2(k+1)]
2
−4(3k+1)=0
⇒4k
2
+4+8k−12k−4=0
⇒4k
2
−4k=0
⇒4k(k−1)=0
⇒k=0 k=1
When k=0
equation y=x
2
+2x+1=0
Roots are x=−1,−1
When k=1
equation 4x
2
+4x+1=0
Roots are x=
2
−1
,
2
−1
Step-by-step explanation:
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