Question

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

• A. 2
• B. 3
• C. 1
• D. 0

Hint:

When dealing with square roots in equations, always check if the solutions satisfy the domain restrictions (e.g., ensuring the terms under the square roots are non-negative).

Answer 1

The Correct Option is C

$\sqrt{(x-1)(x-3)}+\sqrt{(x-3)(x+3)}$
$=\sqrt{4\left(x-\frac{12}{4}\right)\left(x-\frac{2}{4}\right)}$
$\Rightarrow \sqrt{x-3}=0\Rightarrow x=3\text{which is in domain}$
$\text{or}$
$\sqrt{x-1}+\sqrt{x+3}=\sqrt{4x-2}$
$2\sqrt{(x-1)(x+3)}=2x-4$
$x^2+2x-3=x^2-4x+4$
$6x=7$
$x=7/6$(rejected)

Answer 2

Number of Real Roots of the Equation

Given equation:

√(x² - 4x + 3) + √(x² - 9) = √(4x² - 14x + 6)

Step 1: Define the Domain

The square roots are defined when their radicands are non-negative:

x² - 4x + 3 ≥ 0, \quad x² - 9 ≥ 0, \quad 4x² - 14x + 6 ≥ 0

Solving these inequalities:

• x² - 4x + 3 ≥ 0 gives (x - 1)(x - 3) ≥ 0 → x ≤ 1 or x ≥ 3.
• x² - 9 ≥ 0 gives (x - 3)(x + 3) ≥ 0 → x ≤ -3 or x ≥ 3.
• 4x² - 14x + 6 ≥ 0 solves to x ∈ (-∞, 1] ∪ [3, ∞).

Thus, the common domain is x ≤ 1 or x ≥ 3.

Step 2: Squaring Both Sides

Squaring both sides and solving for x, we obtain one valid solution satisfying the given equation.

Final Answer:

The number of real roots is 1.