Math, asked by rehadewan08, 2 months ago

pls solve 13th question

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Answered by prettykitty664
6

\huge\mathtt\red{Answer❣}

qx² - 2px + q = 0

Step-by-step explanation:

If, x = [root(p+q) +root(p-q)] /[root(p+q) - root(p-q)] then find the value of qx2 – 2px + q .

x = (√(p + q)   + √(p-q))/(√(p + q)   - √(p-q))

Multiplying & diving by √(p + q)   + √(p-q)

=> x = (p + q + p - q + 2√(p² - q²) )/ ( (p + q) - (p - q))

=> x  = 2 ( p + √(p² - q²) ) / 2q

=> x = ( p + √(p² - q²) ) / q

=> qx = p + √(p² - q²)

=> qx - p = √(p² - q²)

Squaring both sides

=> q²x² +  p² - 2pqx  = p² - q²

=> q²x² - 2pqx  = - q²

diving by q both sides

=> qx² - 2px = -q

=> qx² - 2px + q = 0

I hope it will help you siso

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Answered by mathdude500
8

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

\rm :\longmapsto\:x = \dfrac{ \sqrt{p + q}  +  \sqrt{p - q} }{\sqrt{p + q} -  \sqrt{p - q} }

\large\underline{\sf{To\:Find - }}

\rm :\longmapsto\:q {x}^{2} - 2px + q

Concept Used :-

Componendo and Dividendo :-

\boxed{ \bf{ If \:   \:  \: \frac{a}{b} =  \frac{c}{d} \:  \implies \:  \frac{a + b}{a - b} =  \frac{c + d}{c - d}}}

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

Given that,

\rm :\longmapsto\:x = \dfrac{ \sqrt{p + q}  +  \sqrt{p - q} }{\sqrt{p + q} -  \sqrt{p - q} }

can be rewritten as

\rm :\longmapsto\:\dfrac{x}{1}  = \dfrac{ \sqrt{p + q}  +  \sqrt{p - q} }{\sqrt{p + q} -  \sqrt{p - q} }

On applying Componendo and Dividendo, we get

\rm :\longmapsto\:\dfrac{x + 1}{x - 1}  = \dfrac{(\sqrt{p + q}+\sqrt{p - q})+ (\sqrt{p + q} - \sqrt{p - q})}{(\sqrt{p + q}  +  \sqrt{p - q} ) - (\sqrt{p + q} -  \sqrt{p - q})}

\rm :\longmapsto\:\dfrac{x + 1}{x - 1}  = \dfrac{\sqrt{p + q}+\sqrt{p - q}+ \sqrt{p + q} - \sqrt{p - q}}{\sqrt{p + q}  +  \sqrt{p - q} - \sqrt{p + q} + \sqrt{p - q}}

\rm :\longmapsto\:\dfrac{x + 1}{x - 1}  = \dfrac{2\sqrt{p + q}}{2\sqrt{p - q}}

\rm :\longmapsto\:\dfrac{x + 1}{x - 1}  = \dfrac{\sqrt{p + q}}{\sqrt{p - q}}

On squaring both sides, we get

\rm :\longmapsto\:\dfrac{ {(x + 1)}^{2} }{ {(x - 1)}^{2} }  = \dfrac{p + q}{p - q}

On applying Componendo and Dividendo, we get

\rm :\longmapsto\:\dfrac{ {(x + 1)}^{2}  +  {(x - 1)}^{2} }{ {(x + 1)}^{2}  -  {(x - 1)}^{2} }  = \dfrac{(p + q) + (p - q)}{(p + q) - (p - q)}

We know,

\boxed{ \rm{  {(x + y)}^{2} +  {(x - y)}^{2}  = 2( {x}^{2}  +  {y}^{2})}}

and

\boxed{ \rm{ {(x + y)}^{2}   -  {(x  -  y)}^{2} = 4xy}}

So, using these Identities, we get

\rm :\longmapsto\:\dfrac{2( {x}^{2}  +  {1}^{2}) }{4x}  = \dfrac{2p}{2q}

\rm :\longmapsto\:\dfrac{{x}^{2}  +  {1}}{2x}  = \dfrac{p}{q}

\rm :\longmapsto\:q {x}^{2} +  q = 2px

\bf :\longmapsto\:q {x}^{2} - 2px + q = 0

Hence, Value of

 \:  \:  \:  \:  \:  \:  \:   \:  \:  \: \purple{\boxed{ \rm{ \bf :\longmapsto\:q {x}^{2} - 2px + q = 0 \:  \:  \:  \:  \: }}}

Additional Information :-

1. Invertendo :-

\boxed{ \bf{ If \:   \:  \: \frac{a}{b} =  \frac{c}{d} \:  \implies \:  \frac{b}{a} =  \frac{d}{c}}}

2. Alternendo

\boxed{ \bf{ If \:   \:  \: \frac{a}{b} =  \frac{c}{d} \:  \implies \:  \frac{a}{c} =  \frac{b}{d}}}

3. Componendo

\boxed{ \bf{ If \:   \:  \: \frac{a}{b} =  \frac{c}{d} \:  \implies \:  \frac{a + b}{ b} =  \frac{c + d}{ d}}}

4. Dividendo

\boxed{ \bf{ If \:   \:  \: \frac{a}{b} =  \frac{c}{d} \:  \implies \:  \frac{a  -  b}{ b} =  \frac{c  -  d}{ d}}}

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