WBJEE · Maths · Quadratic Equation

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 \(a\)
B for exactly one value of \(a\)
C for exactly two values of \(a\)
D for exactly three values of \(a\)

Verified Solution

Answer & Solution

Correct Answer

(B) for exactly one value of \(a\)

Step-by-step Solution

Detailed explanation

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…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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

Step-by-step Solution

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