# 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: 30th Mar 2023

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Answer

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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.

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.

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