UPSC CDS 2 2022 Math Paper

1. For quadratic equation $a x^2+b x+c=0$$x = \;\frac{-b \pm \sqrt{(b^2-4 a c)}}{2 a}$2. $(a+b)(a-b)=a^2-b^2$3. $(a+b)^2=a^2+b^2+2 a b$ Using above formulas to solve $q x^2-2 p x+q=0$By above formula,$x = \;\frac{-b \pm \sqrt{(b^2-4 a c)}}{2 a}$Here, $a=q, b=-2 p$ and $c=q$$x = \;\frac{2 p \pm \sqrt{4 p^2 - 4 q^2}}{2 q}$$\Rightarrow x = \;\frac{p \pm \sqrt{p^2 - q^2}}{q} \; \; \; \cdot \cdot \cdot \cdot \cdot \cdot \cdot (i)$ Statement: 1$x = \;\frac{\sqrt{p+q}+\sqrt{p-q}}{\sqrt{p+q}-\sqrt{p-q}}$, where $p>q$Rationalizing the denominator$\Rightarrow \; \frac{\sqrt{p+q}+\sqrt{p-q}}{\sqrt{p+q}-\sqrt{p-q}} \times \; \frac{\sqrt{p+q}+\sqrt{p-q}}{\sqrt{p+q}+\sqrt{p-q}}$Using the identities (2) \& (3)$\Rightarrow x = \;\frac{p+q+p-q+2 \sqrt{(p+q)(p-q)}}{p+q+p-q}$$\Rightarrow x = \;\frac{p+q+p-q+2 \sqrt{p^2-q^2}}{p+q+p-q}$$\Rightarrow x = \;\frac{p \pm \sqrt{p^2-q^2}}{q}$Hence, statement 1 is correct Statement: 2From equation (2), we can say that,$x = \;\frac{p+\sqrt{\overline{p^2-q^2}}}{q}$Hence, statement 2 is also correct.Statement 3:We have, $x = \;\frac{p \pm \sqrt{\overline{p^2-q^2}}}{q}$For any value of $p \& q$$x = \;\frac{\sqrt{p+q}}{\sqrt{p-q}}$ is not possibleHence, statement 3 is wrong