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Question

Find the value of 'p' for which the quadratic equatio has equal roots, (p+1)n2+2(p+3)n+(p+8)=0

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Solution

We know that while finding the root of a quadratic equation ax2+bx+c=0 by quadratic formula x=b±b24ac2a,
if b24ac>0, then the roots are real and distinct
if b24ac=0, then the roots are real and equal and
if b24ac<0, then the roots are imaginary.

Here, the given quadratic equation (p+1)n2+2(p+3)n+(p+8)=0 is in the form ax2+bx+c=0 where a=(p+1),b=2(p+3)=(2p+6) and c=(p+8).
It is given that the roots are equal, therefore b24ac=0 that is:

b24ac=0(2p+6)2(4×(p+1)×(p+8))=0(2p)2+62+(2×2p×6)4[p(p+8)+1(p+8)]=0(4p2+36+24p)4(p2+8p+p+8)=0(4p2+36+24p)4(p2+9p+8)=0(4p2+36+24p)4p236p32=04p2+36+24p4p236p32=012p+4=012p=4p=412p=13

Hence, p=13

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