If x is a solution of the equation 2 x + 1 - 2 x - 1 = 1 , x ≥ 1 2 , then 4 x 2 - 1 is equal to :

2 x + 1 - 2 x - 1 = 1 . . . 1 ⇒ 2 x + 1 + 2 x - 1 - 2 4 x 2 - 1 = 1 ⇒ 4 x - 1 = 2 4 x 2 - 1 ⇒ 16 x 2 - 8 x + 1 = 16 x 2 - 4 ⇒ 8 x = 5 ⇒ x = 5 8 which satisfies equation 1 . So, 4 x 2 - 1 = 3 4

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Mathematics - Quadratic Equation Question with Solution | TestHub

MathematicsQuadratic EquationTheory of equationsMedium2 minPYQ_2016

MathematicsMediumsingle choice

If $x$ is a solution of the equation $\sqrt{2 x + 1} - \sqrt{2 x - 1} = 1, (x \geq \frac{1}{2})$ , then $\sqrt{4 x^{2} - 1}$ is equal to :

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Answer:

Solution:

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

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

$\Rightarrow 4 x - 1 = 2 \sqrt{4 x^{2} - 1}$

$\Rightarrow 16 x^{2} - 8 x + 1 = 16 x^{2} - 4$

$\Rightarrow 8 x = 5$

$\Rightarrow x = \frac{5}{8}$ which satisfies equation $(1)$ .

So, $\sqrt{4 x^{2} - 1} = \frac{3}{4}$

Stream:JEESubject:MathematicsTopic:Quadratic EquationSubtopic:Theory of equations

⏱ 2mℹ️ Source: PYQ_2016

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