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. 

Answer 1

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`