JEE MainMathematics

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Complex Numbers and Quadratic Equations: Mathematics | JEE Main

The number of real roots of the equation

\sqrt{x^2 - 4x + 3} + \sqrt{x^2 - 9} = \sqrt{4x^2 - 14x + 6}

, is :

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A3B1C2D0

Step-by-step solutionView

Correct answer

The equation has exactly 1 real root, namely $x = 3$ .

Option analysis

Why each option works or fails

A · 3

Squaring the equation repeatedly without factoring creates extraneous roots of higher degree polynomials. Factor the quadratic expressions first to look for common terms before attempting to square radical equations.

B · 1

None. Factoring yields $(x-3)$ common to all radicands, giving $x=3$ as a solution; the remaining equation has no roots in the valid domain. This is the correct option.

C · 2

Finding the algebraic candidate $x = -1/2$ by squaring both sides after dividing by $\sqrt{x-3}$ without verifying it against the domain $x \ge 3$ or $x \le -3$ . Always test candidate solutions in the original radicands to verify that every term inside a square root is non-negative.

D · 0

Assuming that because the remaining reduced equation has no valid roots, the whole equation has no solutions, ignoring $x=3$ . Remember that the factor extracted to simplify the equation, here $\sqrt{|x-3|}$ , provides a root when it equals zero.

Reviewed route

Solution

StepWorking

01given

Given equation: $\sqrt{x^2 - 4x + 3} + \sqrt{x^2 - 9} = \sqrt{4x^2 - 14x + 6}$ .

02goal

Find the number of real solutions $x$ satisfying the equation.

03approach

Factorize the quadratic expressions inside the square roots to identify common factors, extract $\sqrt{x-3}$ , and solve the resulting simpler equations while strictly checking the domain of definition.

04execute

Factor each quadratic under the radical: $x^2 - 4x + 3 = (x-3)(x-1)$ , $x^2 - 9 = (x-3)(x+3)$ , $4x^2 - 14x + 6 = 2(2x^2 - 7x + 3) = 2(2x-1)(x-3) = (4x-2)(x-3)$ . Rewrite the equation: $\sqrt{(x-3)(x-1)} + \sqrt{(x-3)(x+3)} = \sqrt{(4x-2)(x-3)}$ .

05execute

Notice that $x = 3$ makes each radicand zero ( $0 + 0 = 0$ ), so $x = 3$ is a valid real root. For $x > 3$ , we can divide both sides by $\sqrt{x-3}$ : $\sqrt{x-1} + \sqrt{x+3} = \sqrt{4x-2}$ .

06execute

Square both sides of $\sqrt{x-1} + \sqrt{x+3} = \sqrt{4x-2}$ : $(x-1) + (x+3) + 2\sqrt{(x-1)(x+3)} = 4x - 2$ $2x + 2 + 2\sqrt{(x-1)(x+3)} = 4x - 2$ $2\sqrt{(x-1)(x+3)} = 2x - 4$ $\sqrt{x^2 + 2x - 3} = x - 2$ . Squaring both sides again: $x^2 + 2x - 3 = x^2 - 4x + 4$ $6x = 7 \implies x = \frac{7}{6}$ .

✓verify

Check the validity of $x = 7/6$ . For the original terms to be real, we require $x^2 - 9 \ge 0 \implies x \in (-\infty, -3] \cup [3, \infty)$ . Since $7/6 \in (-3, 3)$ , $\sqrt{(7/6)^2 - 9}$ is not real. Also, $\sqrt{x^2+2x-3} = x-2$ requires $x - 2 \ge 0 \implies x \ge 2$ . The value $x = 7/6$ violates this condition since $7/6 < 2$ . Thus, $x = 7/6$ is extraneous. Only $x = 3$ is a real solution. The total number of real roots is 1.

Hints that build this answer step by step

What linear factor is shared by all three quadratic radicands: $x^2 - 4x + 3$ , $x^2 - 9$ , and $4x^2 - 14x + 6$ ?
$(x - 3)$
Evaluating at $x = 3$ gives $0 + 0 = 0$ , confirming $x = 3$ is a solution. What equation remains for $x > 3$ after dividing both sides by $\sqrt{x - 3}$ ?
$\sqrt{x - 1} + \sqrt{x + 3} = \sqrt{4x - 2}$
Squaring both sides of $\sqrt{x - 1} + \sqrt{x + 3} = \sqrt{4x - 2}$ leads to $2\sqrt{x^2 + 2x - 3} = 2x - 4$ . Does this have any valid solutions for $x > 3$ or $x \le -3$ ?
No, squaring again yields $x^2 + 2x - 3 = x^2 - 4x + 4 \implies 6x = 7 \implies x = 7/6$ , which fails the condition $x \ge 2$ and the initial domain.

✓ Source and academic review

Question type: Single correct
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

Why is x = 7/6 rejected even though squaring produced it algebraically?

Squaring can introduce extraneous roots. Additionally, for the square roots in the original equation to exist in the real numbers, each term under the square root must be non-negative. For $x = 7/6$ , $x^2 - 9 < 0$ , making the expressions non-real.

Answer

The equation has exactly 1 real root, namely $x = 3$ .

Why each option works or fails

A: 3 - Squaring the equation repeatedly without factoring creates extraneous roots of higher degree polynomials. Factor the quadratic expressions first to look for common terms before attempting to square radical equations.
B · correct: 1 - None. Factoring yields $(x-3)$ common to all radicands, giving $x=3$ as a solution; the remaining equation has no roots in the valid domain. This is the correct option.
C: 2 - Finding the algebraic candidate $x = -1/2$ by squaring both sides after dividing by $\sqrt{x-3}$ without verifying it against the domain $x \ge 3$ or $x \le -3$ . Always test candidate solutions in the original radicands to verify that every term inside a square root is non-negative.
D: 0 - Assuming that because the remaining reduced equation has no valid roots, the whole equation has no solutions, ignoring $x=3$ . Remember that the factor extracted to simplify the equation, here $\sqrt{|x-3|}$ , provides a root when it equals zero.

Step-by-step solution

given: Given equation: $\sqrt{x^2 - 4x + 3} + \sqrt{x^2 - 9} = \sqrt{4x^2 - 14x + 6}$ .
goal: Find the number of real solutions $x$ satisfying the equation.
approach: Factorize the quadratic expressions inside the square roots to identify common factors, extract $\sqrt{x-3}$ , and solve the resulting simpler equations while strictly checking the domain of definition.
execute: Factor each quadratic under the radical: $x^2 - 4x + 3 = (x-3)(x-1)$ , $x^2 - 9 = (x-3)(x+3)$ , $4x^2 - 14x + 6 = 2(2x^2 - 7x + 3) = 2(2x-1)(x-3) = (4x-2)(x-3)$ . Rewrite the equation: $\sqrt{(x-3)(x-1)} + \sqrt{(x-3)(x+3)} = \sqrt{(4x-2)(x-3)}$ .
execute: Notice that $x = 3$ makes each radicand zero ( $0 + 0 = 0$ ), so $x = 3$ is a valid real root. For $x > 3$ , we can divide both sides by $\sqrt{x-3}$ : $\sqrt{x-1} + \sqrt{x+3} = \sqrt{4x-2}$ .
execute: Square both sides of $\sqrt{x-1} + \sqrt{x+3} = \sqrt{4x-2}$ : $(x-1) + (x+3) + 2\sqrt{(x-1)(x+3)} = 4x - 2$ $2x + 2 + 2\sqrt{(x-1)(x+3)} = 4x - 2$ $2\sqrt{(x-1)(x+3)} = 2x - 4$ $\sqrt{x^2 + 2x - 3} = x - 2$ . Squaring both sides again: $x^2 + 2x - 3 = x^2 - 4x + 4$ $6x = 7 \implies x = \frac{7}{6}$ .
verify: Check the validity of $x = 7/6$ . For the original terms to be real, we require $x^2 - 9 \ge 0 \implies x \in (-\infty, -3] \cup [3, \infty)$ . Since $7/6 \in (-3, 3)$ , $2x + 2 + 2\sqrt{(x-1)(x+3)} = 4x - 2$ is not real. Also, $\sqrt{x^2+2x-3} = x-2$ requires $x - 2 \ge 0 \implies x \ge 2$ . The value $x = 7/6$ violates this condition since $7/6 < 2$ . Thus, $x = 7/6$ is extraneous. Only $x = 3$ is a real solution. The total number of real roots is 1.

Complex Numbers and Quadratic Equations: Mathematics | JEE Main

What feels right?

The equation has exactly 1 real root, namely $x = 3$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

06execute

✓verify

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