If {x^2} + {y^2} = 25 and xy = 12,then find t | vedclass

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Question

(C) Given equations are ${x^2} + {y^2} = 25$ and $xy = 12$.From the second equation,we have $y = \frac{12}{x}$.Substitute this into the first equation

Vedclass · Accepted Answer

$\{3, 4, -3, -4\}$

Vedclass · Answer

(C) Given equations are ${x^2} + {y^2} = 25$ and $xy = 12$.From the second equation,we have $y = \frac{12}{x}$.Substitute this into the first equation: ${x^2} + {\left( \frac{12}{x} \right)^2} = 25$.This simplifies to ${x^2} + \frac{144}{x^2} = 25$.Multiplying by ${x^2}$,we get ${x^4} + 144 = 25{x^2}$,which rearranges to ${x^4} - 25{x^2} + 144 = 0$.Let $u = {x^2}$,then ${u^2} - 25u + 144 = 0$.Factoring the quadratic equation: $(u - 16)(u - 9) = 0$.So,${x^2} = 16$ or ${x^2} = 9$.Taking the square root,we get $x = \pm 4$ or $x = \pm 3$.Thus,the possible values for $x$ are $\{3, 4, -3, -4\}$.

If ${x^2} + {y^2} = 25$ and $xy = 12$,then find the possible values of $x$.

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