Question

and q ≠0 then prove that qx²-px+q=0​

Answer 1

$\huge{\underline{\bold{Answer:}}}$

$\sf x = \dfrac{ \sqrt{p + 2q} + \sqrt{p - 2q} }{ \sqrt{p + 2q} - \sqrt{p - 2q} }$

So, let's rationalise RHS first,

$\sf \frac{ \sqrt{p + 2q} + \sqrt{p - 2q} }{ \sqrt{p + 2q} - \sqrt{p - 2q} } \times \frac{ \sqrt{p + 2q} + \sqrt{p - 2q} }{ \sqrt{p + 2q} + \sqrt{p - 2q} }$

$\sf \dfrac{ (\sqrt{p + 2q} + \sqrt{p - 2q} ) {}^{2} }{ (\sqrt{p + 2q} ) {}^{2} - ( \sqrt{p - 2q}) {}^{2} }$

$\sf \frac{ (\sqrt{p + 2q}) {}^{2} + ( \sqrt{p - 2q}) {}^{2} + 2(\sqrt{p + 2q})(\sqrt{p - 2q})}{ \cancel p + 2q - \cancel p + 2q}$

$\sf \frac{p + 2q + p - 2q + 2 \sqrt{p + 2q} \times \sqrt{p - 2q} }{4q}$

$\sf \frac{2p + 2 \sqrt{p + 2q} \times \sqrt{p - 2q}}{4q}$

$\sf x = \frac{2(p + \sqrt{p + 2q} \times \sqrt{p - 2q})}{2(2q)} \\ \\ \sf x = \frac{p +\sqrt{p + 2q} \times \sqrt{p - 2q}}{2q} \\ \\ \bf on \: cross \: multiplying : - \\ \\ \sf 2qx = p + \sqrt{p + 2q} \times \sqrt{p - 2q} \\ \\ \sf 2qx - p = \sqrt{p + 2q} \times \sqrt{p - 2q} \\ \\ \bf on \: squaring \: both \: sides : \\ \\ \sf (2qx - p) {}^{2} = (\sqrt{p + 2q} \times \sqrt{p - 2q}) {}^{2}$

$\sf{4q {}^{2} x {}^{2} + p {}^{2} - 4qxp= (p + 2q)(p - 2q)} \\ \\ \sf 4q {}^{2} x {}^{2} + p {}^{2} - 4qxp =p {}^{2} - 4q {}^{2} \\ \\ \sf 4q {}^{2} x {}^{2} + p {}^{2} - 4qxp - p {}^{2} + 4q {}^{2} = 0 \\ \\ \sf 4q {}^{2} x{}^{2} - 4qxp + 4q {}^{2} = 0 \\ \\ \sf 4q(qx {}^{2} - xp + q) = 0 \\ \\ \boxed{\bf qx {}^{2} - xp + q= 0}$

$\huge{\underline{\bold{PROVED.}}}$

Answer 2

Step-by-step explanation:

\huge{\underline{\bold{Answer:}}}

Answer:

\sf x = \dfrac{ \sqrt{p + 2q} + \sqrt{p - 2q} }{ \sqrt{p + 2q} - \sqrt{p - 2q} } x=

p+2q

−

p−2q

p+2q

+

p−2q

So, let's rationalise RHS first,

\sf \frac{ \sqrt{p + 2q} + \sqrt{p - 2q} }{ \sqrt{p + 2q} - \sqrt{p - 2q} } \times \frac{ \sqrt{p + 2q} + \sqrt{p - 2q} }{ \sqrt{p + 2q} + \sqrt{p - 2q} }

p+2q

−

p−2q

p+2q

+

p−2q

×

p+2q

+

p−2q

p+2q

+

p−2q

\sf \dfrac{ (\sqrt{p + 2q} + \sqrt{p - 2q} ) {}^{2} }{ (\sqrt{p + 2q} ) {}^{2} - ( \sqrt{p - 2q}) {}^{2} }

(

p+2q

)

2

−(

p−2q

)

2

(

p+2q

+

p−2q

)

2

\sf \frac{ (\sqrt{p + 2q}) {}^{2} + ( \sqrt{p - 2q}) {}^{2} + 2(\sqrt{p + 2q})(\sqrt{p - 2q})}{ \cancel p + 2q - \cancel p + 2q}

p

+2q−

p

+2q

(

p+2q

)

2

+(

p−2q

)

2

+2(

p+2q

)(

p−2q

)

\sf \frac{p + 2q + p - 2q + 2 \sqrt{p + 2q} \times \sqrt{p - 2q} }{4q}

4q

p+2q+p−2q+2

p+2q

×

p−2q

\sf \frac{2p + 2 \sqrt{p + 2q} \times \sqrt{p - 2q}}{4q}

4q

2p+2

p+2q

×

p−2q

\begin{gathered}\sf x = \frac{2(p + \sqrt{p + 2q} \times \sqrt{p - 2q})}{2(2q)} \\ \\ \sf x = \frac{p +\sqrt{p + 2q} \times \sqrt{p - 2q}}{2q} \\ \\ \bf on \: cross \: multiplying : - \\ \\ \sf 2qx = p + \sqrt{p + 2q} \times \sqrt{p - 2q} \\ \\ \sf 2qx - p = \sqrt{p + 2q} \times \sqrt{p - 2q} \\ \\ \bf on \: squaring \: both \: sides : \\ \\ \sf (2qx - p) {}^{2} = (\sqrt{p + 2q} \times \sqrt{p - 2q}) {}^{2} \end{gathered}

x=

2(2q)

2(p+

p+2q

×

p−2q

)

x=

2q

p+

p+2q

×

p−2q

oncrossmultiplying:−

2qx=p+

p+2q

×

p−2q

2qx−p=

p+2q

×

p−2q

onsquaringbothsides:

(2qx−p)

2

=(

p+2q

×

p−2q

)

2

\begin{gathered} \sf{4q {}^{2} x {}^{2} + p {}^{2} - 4qxp= (p + 2q)(p - 2q)} \\ \\ \sf 4q {}^{2} x {}^{2} + p {}^{2} - 4qxp =p {}^{2} - 4q {}^{2} \\ \\ \sf 4q {}^{2} x {}^{2} + p {}^{2} - 4qxp - p {}^{2} + 4q {}^{2} = 0 \\ \\ \sf 4q {}^{2} x{}^{2} - 4qxp + 4q {}^{2} = 0 \\ \\ \sf 4q(qx {}^{2} - xp + q) = 0 \\ \\ \boxed{\bf qx {}^{2} - xp + q= 0}\end{gathered}

4q

2

x

2

+p

2

−4qxp=(p+2q)(p−2q)

4q

2

x

2

+p

2

−4qxp=p

2

−4q

2

4q

2

x

2

+p

2

−4qxp−p

2

+4q

2

=0

4q

2

x

2

−4qxp+4q

2

=0

4q(qx

2

−xp+q)=0

qx

2

−xp+q=0

\huge{\underline{\bold{PROVED.}}}

PROVED.