Question

Hello guys! I need the ans. urgently
$if 2\sqrt{x} + \sqrt{5x - 4\sqrt{x} } =1$
$then \\show \\that x-1 = 0$

Answer 1

Step-by-step explanation:

We have,

$2 \sqrt{x} + \sqrt{5 - 4 \sqrt{x} } = 1$

$\implies \sqrt{5 - 4 \sqrt{x} } = 1 - 2 \sqrt{x}$

$\implies 5 - 4 \sqrt{x} = (1 - 2 \sqrt{x})^{2}$

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

$\implies 5 - 1= 4x$

$\implies 4x = 4$

$\implies x = 1$

$\sf \pink{ \implies x - 1 = 0}$

Answer 2

$\green{\large\underline{\sf{Given- }}}$

$\rm :\longmapsto\:2\sqrt{x} + \sqrt{5x - 4\sqrt{x} } =1$

$\blue{\large\underline{\sf{To\:show - }}}$

$\rm :\longmapsto\:x - 1 = 0$

$\purple{\large\underline{\sf{Solution-}}}$

Given that,

$\rm :\longmapsto\:2\sqrt{x} + \sqrt{5x - 4\sqrt{x} } =1$

To solve, such type of equations, we have to first Shift one of the square root on other side.

So, above equation can be rewritten as

$\rm :\longmapsto\: \sqrt{5x - 4\sqrt{x} } =1 - 2 \sqrt{x}$

On squaring both sides, we get

$\rm :\longmapsto\: \bigg[{\sqrt{5x - 4\sqrt{x} }} \bigg]^{2} =\bigg[1 - 2 \sqrt{x}\bigg]^{2}$

We know that,

$\red{ \boxed{ \sf{ \: {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy}}}$

So, using this identity,

$\rm :\longmapsto\:5x - 4 \sqrt{x} = {1}^{2} + {(2 \sqrt{x}) }^{2} - 2 \times 1 \times 2 \sqrt{x}$

$\rm :\longmapsto\:5x - \cancel{4 \sqrt{x}} = 1 + 4x - \cancel{4 \sqrt{x}}$

$\rm :\longmapsto\:5x = 1 + 4x$

$\rm :\longmapsto\:5x - 1 - 4x = 0$

$\rm :\longmapsto\:x - 1 = 0$

Hence, Proved

Additional Information :-

More Identities to know :-

↝ (a + b)² = a² + 2ab + b²

↝ (a - b)² = a² - 2ab + b²

↝ a² - b² = (a + b)(a - b)

↝ (a + b)² = (a - b)² + 4ab

↝ (a - b)² = (a + b)² - 4ab

↝ (a + b)² + (a - b)² = 2(a² + b²)

↝ (a + b)³ = a³ + b³ + 3ab(a + b)

↝ (a - b)³ = a³ - b³ - 3ab(a - b)