Question

34. If x=
√p + 2q + √p-2q upon
√p+2q-√p-2q
then show that qx^2 - px + q = 0.​

Answer 1

Answer:

$\displaystyle x=\frac{\sqrt{p+2q}+\sqrt{p-2q}}{\sqrt{p+2q}-\sqrt{p-2q}}\\\\=\frac{(\sqrt{p+2q}+\sqrt{p-2q})^2}{(\sqrt{p+2q}-\sqrt{p-2q})(\sqrt{p+2q}+\sqrt{p-2q})}\\\\=\frac{(p+2q)+(p-2q)+2\sqrt{(p+2q)(p-2q)}}{(p+2q)-(p+2q)}\\\\=\frac{2p+2\sqrt{p^2-4q^2}}{4q}\\\\=\frac{p+\sqrt{p^2-4q^2}}{2q}$

$\displaystyle\Rightarrow 2qx = p+\sqrt{p^2-4q^2}\\\\\Rightarrow 2qx-p = \sqrt{p^2-4q^2}\\\\\Rightarrow (2qx-p)^2=p^2-4q^2 \\\\\Rightarrow 4q^2x^2-4pqx+p^2=p^2-4q^2\\\\\Rightarrow 4q^2x^2-4pqx+4q^2=0\\\\\Rightarrow qx^2-px+q=0$

Hope that helps.

Answer 2

Answer:

In attachment I have answer. ↑↑