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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