JEE MainMathematics

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Complex Numbers and Quadratic Equations: Mathematics | JEE Main

Q: Does subtracting the two equations guarantee that the resulting is a common root?

Yes, any value of satisfying both and must also satisfy . Because is linear here, it uniquely isolates the potential common root.

If the value of real number

\text{a} > 0

for which

x^2 - 5\text{a}x + 1 = 0

and

x^2 - \text{a}x - 5 = 0

have a common real root is

\frac{3}{\sqrt{2\beta}}

then

\beta

is equal to _______.

Your answer stays private

What feels right?

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Step-by-step solutionView

Correct answer

The value of $\beta$ is $13$ .

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01given

Given two quadratic equations: $x^2 - 5ax + 1 = 0$ and $x^2 - ax - 5 = 0$ , where $a > 0$ is a real number. They share a common real root, and $a = \frac{3}{\sqrt{2\beta}}$ .

02goal

Find the value of $\beta$ .

03approach

Let $\alpha$ be the common root. Since both equations have identical leading coefficients ( $x^2$ ), subtracting the second equation from the first eliminates the quadratic term and allows us to express the common root $\alpha$ linearly in terms of $a$ . We then substitute $\alpha$ back into either equation to solve for $a$ .

04execute

Subtracting $(x^2 - ax - 5 = 0)$ from $(x^2 - 5ax + 1 = 0)$ gives: $(-5ax + ax) + (1 - (-5)) = 0 \implies -4ax + 6 = 0 \implies x = \frac{6}{4a} = \frac{3}{2a}$ .

05execute

Substitute $x = \frac{3}{2a}$ into $x^2 - 5ax + 1 = 0$ : $\left(\frac{3}{2a}\right)^2 - 5a\left(\frac{3}{2a}\right) + 1 = 0$ $\frac{9}{4a^2} - \frac{15}{2} + 1 = 0 \implies \frac{9}{4a^2} = \frac{13}{2}$ . Thus, $26a^2 = 9 \implies a^2 = \frac{9}{26}$ . Since $a > 0$ , $a = \frac{3}{\sqrt{26}}$ .

06execute

Compare $a = \frac{3}{\sqrt{26}}$ with $a = \frac{3}{\sqrt{2\beta}}$ : $2\beta = 26 \implies \beta = 13$ .

✓verify

Check substitution into the second equation $x^2 - ax - 5 = 0$ : $\frac{9}{4a^2} - a\left(\frac{3}{2a}\right) - 5 = \frac{9}{4(9/26)} - \frac{3}{2} - 5 = \frac{26}{4} - \frac{13}{2} = \frac{13}{2} - \frac{13}{2} = 0$ . Verified.

Hints that build this answer step by step

Let $\alpha$ be the common real root of $x^2 - 5ax + 1 = 0$ and $x^2 - ax - 5 = 0$ . Which algebraic operation eliminates $\alpha^2$ to express $\alpha$ directly in terms of $a$ ?
Subtract the two equations: $(x^2 - 5ax + 1) - (x^2 - ax - 5) = 0$
Solving $-4a\alpha + 6 = 0$ gives which expression for the common root $\alpha$ ?
$\alpha = \frac{3}{2a}$
Substitute $\alpha = \frac{3}{2a}$ into $x^2 - ax - 5 = 0$ . What equation in $a^2$ results?
$\frac{9}{4a^2} - \frac{13}{2} = 0$
From $\frac{9}{4a^2} = \frac{13}{2}$ , solve for $a > 0$ and equate it to $\frac{3}{\sqrt{2\beta}}$ to determine $\beta$ . What is $\beta$ ?
$\beta = 13$

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Does subtracting the two equations guarantee that the resulting

x

is a common root?

Yes, any value of $x$ satisfying both $f(x)=0$ and $g(x)=0$ must also satisfy $f(x) - g(x) = 0$ . Because $f(x) - g(x)$ is linear here, it uniquely isolates the potential common root.

Answer

The value of $\beta$ is $13$ .

Why each option works or fails

Step-by-step solution

given: Given two quadratic equations: $x^2 - 5ax + 1 = 0$ and $x^2 - ax - 5 = 0$ , where $a > 0$ is a real number. They share a common real root, and $a = \frac{3}{\sqrt{2\beta}}$ .
goal: Find the value of $\beta$ .
approach: Let $\alpha$ be the common root. Since both equations have identical leading coefficients ( $x^2$ ), subtracting the second equation from the first eliminates the quadratic term and allows us to express the common root $\alpha$ linearly in terms of $a$ . We then substitute $\alpha$ back into either equation to solve for $a$ .
execute: Subtracting $(x^2 - ax - 5 = 0)$ from $x^2 - ax - 5 = 0$ gives: $(-5ax + ax) + (1 - (-5)) = 0 \implies -4ax + 6 = 0 \implies x = \frac{6}{4a} = \frac{3}{2a}$ .
execute: Substitute $x = \frac{3}{2a}$ into $x^2 - 5ax + 1 = 0$ : $\left(\frac{3}{2a}\right)^2 - 5a\left(\frac{3}{2a}\right) + 1 = 0$ $\frac{9}{4a^2} - \frac{15}{2} + 1 = 0 \implies \frac{9}{4a^2} = \frac{13}{2}$ . Thus, $26a^2 = 9 \implies a^2 = \frac{9}{26}$ . Since $a > 0$ , $a = \frac{3}{\sqrt{26}}$ .
execute: Compare $a = \frac{3}{\sqrt{26}}$ with $a = \frac{3}{\sqrt{2\beta}}$ : $2\beta = 26 \implies \beta = 13$ .
verify: Check substitution into the second equation $x^2 - ax - 5 = 0$ : $\frac{9}{4a^2} - a\left(\frac{3}{2a}\right) - 5 = \frac{9}{4(9/26)} - \frac{3}{2} - 5 = \frac{26}{4} - \frac{13}{2} = \frac{13}{2} - \frac{13}{2} = 0$ . Verified.

Complex Numbers and Quadratic Equations: Mathematics | JEE Main

What feels right?

The value of $\beta$ is $13$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

06execute

✓verify

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Complex Numbers and Quadratic Equations: Mathematics | JEE Main

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