Question

find the value of p for which the quadratic equation
$(p + 1) {x}^{2} - 6(p + 1)x + 3(p + q) = 0$
have equal roots, hence find the roots of the equation. ​

Answer 1

the only thing we need a little while ago but never

Answer 2

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=>(P+1)×2-6(P+1)X+3(P+Q)

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[SOLUTION]:-

Equal roots means,b2=4ac(in quadratic a×2+bx+c=0)

Hence root is X = b2 a

{ APPLYING TO ABOVE }

And to find P, we use b2 = 4ac

36(P+1)2=4×(P+1)2(P+1)=3

3(P+1)=P+Q

2P = Q-3

P = Q-3

HENCE, P=3 & Q= 9