The two equations x3+1=0 and ax2+bx+c=0,a,b,c∈R have two roots in common. Then a+b is equal to
A
2
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B
0
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C
−1
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D
3
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Solution
The correct option is B0 x3+1=0⇒(x+1)(x2−x+1)=0 x2−x+1=0 has non-real roots as Δ<0
We know that, for a quadratic equation with real coefficients, imaginary roots always occur in conjugate pair. ∴x2−x+1=0 and ax2+bx+c=0 have both roots in common. ⇒1a=−1b=1c ⇒a+b=0