Question

If x = -1 is a common root of both the equations \(2x^2 + 3x + p = 0\) and \(qx^2 - qx + 4 = 0\) then the value of \(p + q\) is

(A) 1

(B) -1

(C) 2

(D) -2

Answer 1

Correct option is (B) -1

Given,

-1 is root of equation \(2 x^{2}+3 x+p=0\)

Put \(x=-1\) in the equation \(2 x^{2}+3 x+p=0\)

\(2(-1)^{2}+3(-1)+P =0\)

\(2 \times 1-3+P =0\)

\(2-3+P =0\)

\(-1+P=0\)

\(\therefore P =1\)

Given,

Second equation is \(q x^{2}-q x+4=0\)

Put \(x=-1\) in the equation \(q x^{2}-q x+4=0\)

\(q(-1)^{2}-q(-1)+4 =0 \)

\(q+q+4 =0\)

\(2 q+4 =0\)

\(2q =-4 \)

\(\therefore q =\frac{-4}{2}=-2\)

Therefore,

\(p+q\)

\(= 1+(-2)\)

\(= 1-2\)

\(= -1\)