Question

help me solve this from the chapter quadratic equations class 10th​

Answer 1

Question :-

$\rm{ \sqrt{ \frac{x}{1 - x} } - \sqrt{ \frac{1 - x}{x} } = \frac{3}{2} }$

Solution:-

$\sf:\longmapsto{ \frac{( \sqrt{x} \times \sqrt{x} ) - ( \sqrt{1 - x} \times \sqrt{1 - x} ) }{ \sqrt{ x(1 - x)}} = \frac{3}{2} } \\ \\ \sf:\longmapsto{ \frac{x - (1 - x)}{ \sqrt{ x(1 - x)}} = \frac{3}{2} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf:\longmapsto{ \frac{x - 1 + x}{ \sqrt{ x(1 - x)}} = \frac{3}{2} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf:\longmapsto{ \frac{2x - 1}{ \sqrt{ x(1 - x)}} = \frac{3}{2} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf:\longmapsto{2(2x - 1) = 3 [\sqrt{x(1 - x)}]</p><p>} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf:\longmapsto{4x - 2 = 3 [\sqrt{x(1 - x)}]</p><p>} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Squaring both sides

$\sf:\longmapsto{4x {}^{2} - 4 = 9 [x (1 - x)]} \: \: \: \: \: \: \\ \\ \sf:\longmapsto{4x {}^{2} - 4 =9(x - x {}^{2} ) } \: \: \: \: \: \: \: \: \\ \\ \sf:\longmapsto{4x {}^{2} - 4 = 9x - 9x {}^{2} } \: \: \: \: \: \: \: \: \: \\ \\ \sf:\longmapsto{9x {}^{2} + 4x {}^{2} - 9x - 4 = 0} \\ \\ \sf:\longmapsto{13x {}^{2} - 9x - 4 = 0} \: \: \: \: \: \: \: \: \:$

In Quadratic equation we easy to solve multiply a × c , where the general form of Quadratic equation is ax² + bx + c = 0.

$\sf:\longmapsto{x {}^{2} - 9x - 52 }$

By splinting the middle term

$\sf:\longmapsto{x {}^{2} - 13x + 4x - 52 = 0} \: \: \: \: \: \\ \\ \sf:\longmapsto{x(x - 13) + 4(x - 13) = 0} \\ \\ \sf:\longmapsto{(x - 13)(x + 4) = 0} \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf:\longmapsto{x - 13 = 0} \\ \bf:\longmapsto \red{x = 13} \: \: \: \: \: \: \\ \\ \sf:\longmapsto{x + 4 = 0} \\ \bf:\longmapsto \red{x = - 4} \: \: \: \:$

Answer :

• x = 13
• x = -4

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Answer 2

$\large\underline{\sf{Given \:Question - }}$

Solve for x :-

$\rm :\longmapsto\: \sqrt{\dfrac{x}{1 - x} } - \sqrt{\dfrac{1 - x}{x} } = \dfrac{3}{2}$

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

Given equation is

$\rm :\longmapsto\: \sqrt{\dfrac{x}{1 - x} } - \sqrt{\dfrac{1 - x}{x} } = \dfrac{3}{2}$

Let we assume that,

$\red{\rm :\longmapsto\: \sqrt{\dfrac{x}{1 - x} } = y - - - - (1)}$

So, given equation can be rewritten as

$\rm :\longmapsto\:y - \dfrac{1}{y} = \dfrac{3}{2}$

$\rm :\longmapsto\:\dfrac{ {y}^{2} -1}{y} = \dfrac{3}{2}$

$\rm :\longmapsto\:2( {y}^{2} - 1) = 3y$

$\rm :\longmapsto\:2{y}^{2} -2 = 3y$

$\rm :\longmapsto\:2{y}^{2} -2 - 3y = 0$

$\rm :\longmapsto\:2{y}^{2} - 3y - 2 = 0$

$\rm :\longmapsto\:2{y}^{2} - 4y + y - 2 = 0$

$\rm :\longmapsto\:2y(y - 2) + 1(y - 2) = 0$

$\rm :\longmapsto\:(y - 2)(2y + 1) = 0$

$\bf\implies \:y = 1 \: \: \: or \: \: \: y = - \: \dfrac{1}{2}$

So, When

$\rm :\longmapsto\:y = 1$

$\rm :\longmapsto\: \sqrt{\dfrac{x}{1 - x} } = 1$

On squaring both sides, we get

$\rm :\longmapsto\:\dfrac{x}{1 - x} = 1$

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

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

$\rm :\longmapsto\:2x= 1$

$\bf\implies \:x = \dfrac{1}{2}$

Now, when

$\rm :\longmapsto\:y = - \: \dfrac{1}{2}$

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

On squaring both sides, we get

$\rm :\longmapsto\:\dfrac{x}{1 - x} = \dfrac{1}{4}$

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

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

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

$\bf\implies \:x = \dfrac{1}{5}$

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:Solution \: is \: -\begin{cases} &\sf{x = \dfrac{1}{2} } \\ \\ &\sf{x = \dfrac{1}{5} } \end{cases}\end{gathered}\end{gathered}$

Additional Information :-

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

If Discriminant, D > 0, then roots of the equation are real and unequal.

If Discriminant, D = 0, then roots of the equation are real and equal.

If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

Discriminant, D = b² - 4ac