What is the one of the roots of the equation -2x3 - x - -3 - x2x - 32-

The correct option is B 2 √(2x3 x) √(3 x2x) = 32 (Let t = √(2x3 x)) ⇒ t 1t = 32⇒ 2t^2 3t 2 = 0⇒ (t 2)(2t+1) = 0 ⇒t = 2, 12⇒√(2x3 x) = ...

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Standard XII

Mathematics

Discriminant

What is the o...

Question

What is the one of the roots of the equation $\sqrt{\frac{2 x}{3 - x}} - \sqrt{\frac{3 - x}{2 x}} = \frac{3}{2}$ ?

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Solution

The correct option is B $2$
$\sqrt{\frac{2 x}{3 - x}} - \sqrt{\frac{3 - x}{2 x}} = \frac{3}{2}$ (Let $t = \sqrt{\frac{2 x}{3 - x}})$
$\Rightarrow t - \frac{1}{t} = \frac{3}{2} \Rightarrow 2 t^{2} - 3 t - 2 = 0 \Rightarrow (t - 2) (2 t + 1) = 0$
$\Rightarrow t = 2, - \frac{1}{2} \Rightarrow \sqrt{\frac{2 x}{3 - x}} = 2, \frac{- 1}{2} \Rightarrow \frac{2 x}{3 - x} = 4, \frac{1}{4}$
$\Rightarrow 2 x = 12 - 4 x \Rightarrow x = 2$ & $x = \frac{1}{3}$

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What is the one of the roots of the equation √2x3−x−√3−x2x=32?

The correct option is B 2√2x3−x−√3−x2x=32 (Let t=√2x3−x)⇒t−1t=32⇒2t2−3t−2=0⇒(t−2)(2t+1)=0⇒t=2,−12⇒√2x3−x=2,−12⇒2x3−x=4,14⇒2x=12−4x⇒x=2 & x=13

What is the one of the roots of the equation $\sqrt{\frac{2 x}{3 - x}} - \sqrt{\frac{3 - x}{2 x}} = \frac{3}{2}$ ?