Question

If x = √p+2q+√p-2q/√p+2q-√p-2q then show that qx2 -px+q=0

Answer 1

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Answer 2

From given $x=\frac{\sqrt{p+2q}+\sqrt{p-2q}}{\sqrt{p+2q}-\sqrt{p-2q}}$ we have proved that $qx^2-px+q=0$

Step-by-step explanation:

• Given that $x=\frac{\sqrt{p+2q}+\sqrt{p-2q}}{\sqrt{p+2q}-\sqrt{p-2q}}$
• To prove that $qx^2-px+q=0$
• $x=\frac{\sqrt{p+2q}+\sqrt{p-2q}}{\sqrt{p+2q}-\sqrt{p-2q}}$
• Multiply and dividing by the conjugate $\sqrt{p+2q}+\sqrt{p-2q}$ we get

• $x=\frac{\sqrt{p+2q}+\sqrt{p-2q}}{\sqrt{p+2q}-\sqrt{p-2q}}\times \frac{\sqrt{p+2q}+\sqrt{p-2q}}{\sqrt{p+2q}+\sqrt{p-2q}}$

• $x=\frac{(\sqrt{p+2q}+\sqrt{p-2q})^2}{\sqrt{p+2q}^2-\sqrt{p-2q}^2}$ ( using the properties $(a+b)^2=a^2+b^2+2ab$ and  $(a-b)(a+b)=a^2-b^2$ )

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

• $x=\frac{p+2q+p-2q+2\sqrt{(p+2q)(p-2q)}}{p+2q-p+2q}$
• $x=\frac{2p+2\sqrt{p^2-(2q)^2}}{4q}$ ( using the property $(a-b)(a+b)=a^2-b^2$ )

• $x=2(\frac{p+\sqrt{p^2-4q^2}}{4q})$
• $x=\frac{p+\sqrt{p^2-4q^2}}{2q}$
• $(2q)x=p+\sqrt{p^2-4q^2}$
• $2qx-p=\sqrt{p^2-4q^2}$
• Now squaring on both sides we get
• $(2qx-p)^2=(\sqrt{p^2-4q^2})^2$
• $(2qx)^2+p^2-2(2qx)(p)=p^2-4q^2$

• $4q^2x^2-4qpx=-4q^2$
• Dividing by 4q on both sides we get
• $\frac{4q^2x^2-4qpx}{4q}=-\frac{4q^2}{4q}$
• $qx^2-px=-q$

• Therefore $qx^2-px+q=0$

Hence proved