Question

Please solve it
Std 10 ( ratio and proportion)

Answer 1

$\LARGE{ \underline{\underline{ \orange{ \sf{Required \: answer:}}}}}$

GiveN:

• $\rm{ \dfrac{ \sqrt{a + x} + \sqrt{a - x} }{ \sqrt{a + x} - \sqrt{a - x} } = b}$

To FinD:

• We have to solve for x.

Step-wise-Step Explanation:

Multiply $\rm{ \sqrt{a + x} + \sqrt{a - x} }$ in both Numerator and denominator in LHS.

⇒ $\rm{ \dfrac{ (\sqrt{a + x} + \sqrt{a - x} )( \sqrt{a + x} + \sqrt{a - x} )}{ (\sqrt{a + x} - \sqrt{a - x})( \sqrt{a + x} + \sqrt{a - x} )} = b}$

Simplifying using Identities,

⇒ $\rm{ \dfrac{( \sqrt{a + x} + \sqrt{a - x} ) {}^{2} }{( \sqrt{a + x} ) {}^{2} - ( \sqrt{a - x} ) {}^{2} } = b}$

⇒ $\rm{ \dfrac{a + x + a - x + 2 \sqrt{ {a}^{2} - {x}^{2} } }{a + x - a + x} = b}$

⇒ $\rm{ \dfrac{2a + 2 \sqrt{ {a}^{2} - {x}^{2} } }{2x} = b}$

Taking 2 in common from numerator and denominator and cancelling it.

⇒ $\rm{ \dfrac{a + \sqrt{ {a}^{2} - {x}^{2} } }{x} = b}$

Now multiplying x in RHS because inverse of division is multiplication.

⇒ $\rm{a + \sqrt{ {a}^{2} - {x}^{2} } = bx}$

Subtracting a from RHS,

⇒ $\rm{ \sqrt{ {a}^{2} - {x}^{2} } = bx - a}$

Squaring both sides,

⇒ $\rm{ {a}^{2} - {x}^{2} = {b}^{2} {x}^{2} + {a}^{2} - 2abx}$

⇒ $\rm{ {b}^{2} {x}^{2} + {x}^{2} - 2abx = 0}$

Taking x common, then it will be 0.

⇒ $\rm{ {b}^{2} x + x - 2ab = 0}$

Now taking x common from b²x and x.

⇒ $\rm{x( {b}^{2} + 1) - 2ab = 0}$

Then, x will be equal to:

⇒ $\rm{ \red{x = \dfrac{2ab}{ {b}^{2} + 1 } }}$

And hence this is the required value of x.

Answer 2

♣ Qᴜᴇꜱᴛɪᴏɴ :

Using the properties of proportion :

Solve for x : $\sf{\displaystyle\frac{\sqrt{a+x}+\sqrt{a-x}}{\sqrt{a+x}-\sqrt{a-x}}=b}$

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♣ ᴀɴꜱᴡᴇʀ :

$\large\boxed{\sf{x=\dfrac{2ab}{b^2+1}}}$

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♣ ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴꜱ :

$\sf{\dfrac{\sqrt{a+x}+\sqrt{a-x}}{\sqrt{a+x}-\sqrt{a-x}}=b}$

Solve it using Componendo and Dividendo Rule :

This rule states :

If $\sf{\dfrac{a}{b}=\dfrac{c}{d}}$ , then $\sf{\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}}$

$\sf{\dfrac{\sqrt{a+x}+\sqrt{a-x}}{\sqrt{a+x}-\sqrt{a-x}}=\dfrac{b}{1}}$

$\implies\sf{\dfrac{(\sqrt{a+x}+\sqrt{a-x})+(\sqrt{a+x}-\sqrt{a-x})}{(\sqrt{a+x}+\sqrt{a-x})-(\sqrt{a+x}-\sqrt{a-x})}=\dfrac{b+1}{b-1}}$

$\implies\sf{\dfrac{\sqrt{a+x}+\sqrt{a-x}+\sqrt{a+x}-\sqrt{a-x}}{\sqrt{a+x}+\sqrt{a-x}-\sqrt{a+x}+\sqrt{a-x}}=\dfrac{b+1}{b-1}}$

$\implies\sf{\dfrac{\sqrt{a+x}+\sqrt{a+x}}{\sqrt{a-x}+\sqrt{a-x}}=\dfrac{b+1}{b-1}}$

$\implies\sf{\dfrac{2\cdot\sqrt{a+x}}{2\cdot\sqrt{a-x}}=\dfrac{b+1}{b-1}}$

$\implies\sf{\dfrac{\sqrt{a+x}}{\sqrt{a-x}}=\dfrac{b+1}{b-1}}$

Squaring we get,

$\sf{\left(\dfrac{\sqrt{a+x}}{\sqrt{a-x}}\right)^2=\left(\dfrac{b+1}{b-1}\right)^2}$

$\implies\sf{\dfrac{(\sqrt{a+x})^2}{(\sqrt{a-x})^2}=\dfrac{(b+1)^2}{(b-1)^2}}$

$\implies\sf{\dfrac{a+x}{a-x}=\dfrac{b^2+2b+1}{b^2-2b+1}}$

Apply Componendo and Dividendo Rule again :

If $\sf{\dfrac{a}{b}=\dfrac{c}{d}}$ , then $\sf{\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}}$

$\implies\sf{\dfrac{a+x}{a-x}=\dfrac{b^2+2b+1}{b^2-2b+1}}$

$\implies\sf{\displaystyle\frac{\left(a+x\right)+\left(a-x\right)}{\left(a+x\right)-\left(a-x\right)}=\frac{\left(b^2+2b+1\right)+\left(b^2-2b+1\right)}{\left(b^2+2b+1\right)-\left(b^2-2b+1\right)}}$

$\implies\sf{\dfrac{a+x+a-x}{a+x-a+x}=\dfrac{b^2+2b+1+b^2-2b+1}{b^2+2b+1-b^2+2b-1}}$

$\implies\sf{\dfrac{a+a}{x+x}=\dfrac{b^2+b^2+2}{2b+2b}}$

$\implies\sf{\dfrac{2a}{2x}=\dfrac{2b^2+2}{4b}}$

$\implies\sf{\dfrac{a}{x}=\dfrac{2b^2+2}{4b}}$

$\implies\sf{\dfrac{x}{a}=\dfrac{4b}{2b^2+2}}$

Multiplying both sides by a :

$\implies\sf{\dfrac{x}{a} \times \:a=\dfrac{4b}{2b^2+2}\times\:a}$

$\implies\sf{x=\dfrac{4ab}{2b^2+2}}$

Taking 2 as common in R.H.S :

$\implies\sf{x=\dfrac{2(2ab)}{2(b^2+1)}}$

Cancelling 2 in numerator and denominator :

$\large\boxed{\sf{x=\dfrac{2ab}{b^2+1}}}$