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

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(C) Given: ${x^2} + {y^2} = 25$ and ${xy} = 12$.From the second equation,$y = \frac{12}{x}$.Substituting this into the first equation: ${x^2} + \left(

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

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(C) Given: ${x^2} + {y^2} = 25$ and ${xy} = 12$.From the second equation,$y = \frac{12}{x}$.Substituting this into the first equation: ${x^2} + \left( \frac{12}{x} \right)^2 = 25$.${x^2} + \frac{144}{x^2} = 25$.Multiplying by ${x^2}$: ${x^4} + 144 = 25{x^2}$.Rearranging: ${x^4} - 25{x^2} + 144 = 0$.Let ${x^2} = t$,then ${t^2} - 25t + 144 = 0$.Factoring: $(t - 16)(t - 9) = 0$.So,${x^2} = 16$ or ${x^2} = 9$.Therefore,$x = \pm 4$ or $x = \pm 3$.The set of possible values for $x$ is $\{3, 4, -3, -4\}$.

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

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