The number of real roots of the equation x ^ 2 - 4x + 3 + x ^ 2 - 9 = 4 x ^ 2 - 14x + 6 , is:

Correct answer: B. 1 Given, x ^ 2 - 4x + 3 + x ^ 2 - 9 = 4 x ^ 2 - 14x + 6 ⇒ (x - 3)(x - 1) + (x - 3)(x + 3) = (x - 3)(4x - 2) ⇒ (x - 3) (x - 1) + (x - 3) (x + 3) - (x - 3) (4x - 2) = 0 ⇒ (x - 3) = 0or[ (x - 1) + (x + 3) - (4x - 2) ] = 0 So, x = 3 is one solution and also x ≥ 3 Now solving [ (x - 1) + (x + 3) - (4x - 2) ] = 0 ⇒ (x - 1) + (x + 3) = (4x - 2) Now squaring both side we get, ⇒ (x - 1) + (x + 3) + 2 (x - 1)(x + 3) = (4x - 2) ⇒ 2 (x - 1)(x + 3) = (2x - 4) ⇒ (x - 1)(x + 3) = (x - 2) ^ 2 ⇒ x ^ 2 + 2x - 3 = x ^ 2 - 4x + 4 ⇒ 6x = 7 ⇒ x = (7)/(6) which does not satisfy the given expression and x ≥ 3 So, there is only one solution which is x = 3

JEE Main 202331 JanuaryMorning Shift

Exam: JEE Main
Year: 2023
Date: 31 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Quadratic Equation

JEE Main 2023 · 31 January · Morning Shift · Mathematics · Quadratic Equation

The number of real roots of the equation x ^ 2 - 4x + 3 + x ^ 2 - 9...

The number of real roots of the equation

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

, is:

Options

A $0$
B $1$
C $3$
D $2$

Correct answer

B. $1$

Step-by-step solution

Given,

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

$⇒ \sqrt{\left(x - 3\right)\left(x - 1\right)} + \sqrt{\left(x - 3\right)\left(x + 3\right)} = \sqrt{\left(x - 3\right)\left(4x - 2\right)}$

$⇒ \sqrt{\left(x - 3\right)}\sqrt{\left(x - 1\right)} + \sqrt{\left(x - 3\right)}\sqrt{\left(x + 3\right)} - \sqrt{\left(x - 3\right)}\sqrt{\left(4x - 2\right)} = 0$

$⇒ \sqrt{\left(x - 3\right)} = 0\text{or}\left[\sqrt{\left(x - 1\right)} + \sqrt{\left(x + 3\right)} - \sqrt{\left(4x - 2\right)}\right] = 0$

So, $x = 3$ is one solution and also $x \ge 3$

Now solving \left[\sqrt{\left(x - 1\right)} + \sqrt{\left(x + 3\right)} - \sqrt{\left(4x - 2\right)}\right] = 0 $$ ⇒ \sqrt{\left(x - 1\right)} + \sqrt{\left(x + 3\right)} = \sqrt{\left(4x - 2\right)} $Now squaring both side we get,$ ⇒ \left(x - 1\right) + \left(x + 3\right) + 2\sqrt{\left(x - 1\right)\left(x + 3\right)} = \left(4x - 2\right) $$ ⇒ 2\sqrt{\left(x - 1\right)\left(x + 3\right)} = \left(2x - 4\right) $$ ⇒ \left(x - 1\right)\left(x + 3\right) = {\left(x - 2\right)}^{2} $$ ⇒ {x}^{2} + 2x - 3 = {x}^{2} - 4x + 4 $$ ⇒ 6x = 7 ⇒ x = \dfrac{7}{6} $which does not satisfy the given expression and$ x \ge 3 $So, there is only one solution which is$ x = 3$

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