Q84 / 160 · WBJEE 2012 · Mathematics

The equations $x^{2}+x+a=0 \quad$ and $x^{2}+a x+1=0$ have a common real root

  • A
    for no value of $a$
  • Correct answer
    B
    for exactly one value of $a$
  • C
    for exactly two values of $a$
  • D
    for exactly three values of $a$

Solution

Let \(\alpha\) be the common roots \(\begin{array}{l} \therefore \alpha^{2}+\alpha+a=0 \ldots \text { (i) } \\ \text { and } \alpha^{2}+a \alpha+1=0 \ldots \text { (ii) } \\ \frac{\alpha^{2}}{1-a^{2}}=\frac{\alpha}{a-1}=\frac{1}{a-1}[a \neq 1] \end{array}\) Eliminating \(\alpha\), we get \(\begin{array}{l} (a-1)^{2}=\left(1-a^{2}\right)(a-1) \\ \Rightarrow a-1)=1-a^{2} \\ \Rightarrow a^{2}+a-2=0 \\ \Rightarrow a^{2}+2 a-a-2=0 \\ \Rightarrow a(a+2)-1(a+2)=0 \\ \Rightarrow(a+2)(a-1)=0 \\ \Rightarrow a=-2 \cdot[\because a \neq 1] \end{array}\)

Hence, the correct answer is option: 'B'.

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