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Question
If -4 is a root of the equation `x^2+2x+4p=0` find the value of k for the which the quadratic equation ` x^2+px(1+3k)+7(3+2k)=0` has equal roots.
Solution
It is given that -4 is a root of the quadratic equation` x^2+2x+4p=0`
∴`(-4)^2+2xx(-4)+4p=0`
⇒`16-8+4p=0`
⇒`4p+8=0`
⇒`p=-2`
The equation `x^2+px(1+3k)+7(3+2k)=0` has real roots`
∴` D=0`
⇒ `[p(1+3k)]^2-4xx1xx7(3+2k)=0`
⇒ `[-2(1+3k)]^2-28(3+2k)=0`
⇒`4(1+6k+9k^2)-28(3+2k)=0`
⇒`4(1+6k+9k^2-21-14k)=0`
⇒ `9k^2-8k-20=0`
⇒`9k^2-18k+10(k-2)=0`
⇒`(k-2)(9k+10)=0`
⇒`k-2=0 or 9k+10=0`
⇒ `k=2 or k=-10/9`
Hence, the required value of k is `2 or -10/9`
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