Q20 of 47 Page 1

Find that value of p for which the quadratic equation had equal roots. Hence find the roots of the equation.

(p + 1)x2-6(p + 1)x + 3(p + 9)=0


This is a quadratic equation of form ax2 + bx + c=0


Where a=p + 1, b=-6(p + 1) ,c=3(p + 9)


For equal roots, D=0


D=b2-4ac=0


Substituting the values of a,b and c


(-6(p + 1))2-4(p + 1)3(p + 9)=0


⇒ 36(p2 + 1 + 2p)-12(p2 + 10p + 9)=0


⇒ 12(2p2-4p-6)=0


⇒ p2-2p-3=0


⇒ p2-3p + p-3=0


⇒ p(p-3) + 1(p-3)=0


⇒ (p + 1)(p-3)=0


There are two possible values of p, p=-1 and p=3


We neglect p=-1 as on putting p=-1; the equation does not remain quadratic.


∴ p=3 is our required value of p.


Now,



As we know roots are equal,


Therefore, each root=


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