Question : If $x+y=4, x^{2}+y^{2}=14$ and $x>y$, then the correct value of $x$ and $y$ is:
Option 1: $2+\sqrt{3} \text{ and } 2-\sqrt{3}$
Option 2: $2-\sqrt{2} \text{ and } \sqrt{3}$
Option 3: $3 \text{ and } 1$
Option 4: $2+\sqrt{3} \text{ and } 2\sqrt{2}$
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Correct Answer: $2+\sqrt{3} \text{ and } 2-\sqrt{3}$
Solution : Given: $x^{2}+y^{2}=14$ -----------(i) ⇒ $x+y=4$ ----------------(ii) Squaring on both sides of equation (ii), $(x+y)^{2} = 4^{2}$ ⇒ $x^{2}+y^{2}+2xy = 16$ ⇒ $14+2xy = 16$ ⇒ $2xy = 2$ Subtracting $2xy$ on both sides from equation (i) $x^{2}+y^{2}-2xy = 14-2xy$ ⇒ $(x-y)^{2} = 12$ ⇒ $(x-y) = \sqrt12$ ⇒ $x-y =2\sqrt3$ -----------(iii) From (ii) and (iii) $x = 2+ \sqrt3$ and $y = 2- \sqrt3$ Hence, the correct answer is $2+ \sqrt3 \text{ and } 2- \sqrt3$.
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