Question

Q. The solution of
$\sqrt{5x - 1} + \sqrt{x - 1} = 2$
A. x= 2, x= 1
B. x= 1
C. x=2
D. none

NTSE-II​

Answer 1

Solution:

The given equation is here:

$\sqrt{5x - 1} + \sqrt{x - 1} = 2$

Squaring on both sides:

$5x - 1 = 4 + x - 1 - 4 \sqrt{x - 1}$

$x - 1 = - \sqrt{x - 1}$

Squaring on both sides:

${(x - 1)}^{2} = ({ - \sqrt{x - 1} })^{2}$

$= > {x}^{2} - 2x + 1 = x - 1$

$= > {x}^{2} - 3x + 1 + 1 = 0$

$= > {x}^{2} - 3x + 2 = 0$

$= > (x - 2)(x - 1) = 0$

$Therefore, \: x = 2 \: or \: x = 1$

But x = 2 is not satisfying the given equation.

So, x = 1 is the only solution.

[ Option C. x = 1]

Answer 2

Answer:

Option(B)

Step-by-step explanation:

Given Equation is : √5x - 1 + √x - 1 = 2

⇒ (√5x - 1) = 2 - (√x - 1)

On Squaring both sides, we get

⇒ (√5x - 1)² = [2 - (√x - 1)]²

⇒ 5x - 1 = 4 + (x - 1) - 4(√x - 1)

⇒ 5x - 1 = 4 + x - 1 - 4√x - 1

⇒ 4√x - 1 = -4x + 4

On Squaring both sides, we get

⇒ (4√x - 1)² = (-4x + 4)²

⇒ [16(x - 1)] = 16x² + 16 - 32x

⇒ 16x - 16 = 16x² + 16 - 32x

⇒ 16x² - 48x + 32 = 0

⇒ x² - 3x + 2 = 0

⇒ x² - 2x - x + 2 = 0

⇒ x(x - 2) - (x - 2) = 0

⇒ (x - 1)(x - 2) = 0

⇒ x = 1, 2{It doesn't satisfy the condition}

⇒ x = 1.

Hope it helps!