If the value of real number a 0 for which x^2 - 5ax + 1 = 0 and x… - Solution | JEE Mathematics

NVAMediumJEE 2023Quadratic Equations in Complex Numbers

If the value of real number $a > 0$ for which $x^2 - 5ax + 1 = 0$ and $x^2 - ax - 5 = 0$ have a common real root is $\frac{3}{\sqrt{2}}$ , then $\beta$ is equal to .....

Answer

Correct answer:13

Step-by-step solution

Standard Method

Given: The equations $x^2 - 5ax + 1 = 0$ and $x^2 - ax - 5 = 0$ have a common real root.

Find: $\beta$ .

Two equations have common root.

(4a)(26a) = (-6)^2 = 36

a^2 = \frac{26}{9}

a = \frac{\sqrt{26}}{3}

\beta = 13

Therefore, the required value is $13$ .

Approach Solution - 2

Given: The two equations have a common root.

Find: $\beta$ .

From the extracted solution:

(4a)(26a) = (-6)^2 = 36

This gives

a^2 = \frac{9}{26}

Hence,

a = \frac{3}{\sqrt{26}}

So,

\beta = 13

Therefore, the correct answer is $13$ .

Common mistakes

Using the common-root condition incorrectly. When two quadratics share a root, one must apply the relation derived from elimination carefully; writing the condition with wrong coefficients leads to an incorrect value of $a$ . Recheck coefficient comparison before solving.
Ignoring the condition $a > 0$ . Even if $a^2$ is found correctly, taking the negative square root would violate the given condition. Always choose the positive root here.
Confusing the intermediate value of $a$ with the required value of $\beta$ . The question asks for $\beta$ , not for $a$ , so the final step must be read from the given relation in the solution.

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