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The equation ${x^2} + x + a = 0$ and ${x^2} + ax + 1 = 0$ have a common real root.            E. For no values of a            F. For exactly one value of a            G. For exactly two values of a            H. For exactly three values of a

Last updated date: 07th Sep 2024
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Hint: First check whether the root of the equation satisfies the equation or not and then analyse the answer.
We have given the quadratic equations ${x^2} + x + a = 0{\text{ - - - - - - - - - - }}(1)$ and ${x^2} + ax + 1 = 0{\text{ - - - - - - - - - - - }}(2)$ .Let the common real root is $\alpha$ , then $\alpha$ will satisfy both the equations. From the first equation we’ll get ${\alpha ^2} + \alpha + a = 0{\text{ - - - - - }}(3)$ and from second, ${\alpha ^2} + a\alpha + 1 = 0{\text{ - - - - - }}(4)$ . Now, we have to solve these equations for the value of $\alpha$ .On subtracting them we’ll get,

$\ {\alpha ^2} + \alpha + a = 0{\text{ - - - - - }}(3) \\ {\alpha ^2} + a\alpha + 1 = 0{\text{ - - - - }}(4) \\ \Rightarrow {\alpha ^2} - {\alpha ^2} + \alpha - a\alpha + a - 1 = 0 \\ \Rightarrow {{{\alpha }}^2} - {{{\alpha }}^2} + \alpha - a\alpha + a - 1 = 0 \\ \Rightarrow \alpha (1 - a) + a - 1 = 0 \\ \Rightarrow (1 - a)[\alpha - 1] = 0 \\ \Rightarrow \alpha = 1 \\ \$

On putting this value of $\alpha$ in third equation we’ll get,
$\ 1 + 1 + a = 0 \\ \Rightarrow a = - 2 \\ \$
So, the given condition is true for exactly one value of a.
Hence option B is correct.
Note: Here we need to be clear, what we are actually looking for. In this question, we had to find the value of a given certain condition. We applied that given condition and solved it.