(sqrt{4489}-sqrt{2601})=(?)^{2} | vedclass

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Question

(C) First,calculate the square root of $4489$: $\sqrt{4489} = 67$.Next,calculate the square root of $2601$: $\sqrt{2601} = 51$.Subtract the two values

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) First,calculate the square root of $4489$: $\sqrt{4489} = 67$.Next,calculate the square root of $2601$: $\sqrt{2601} = 51$.Subtract the two values: $67 - 51 = 16$.Given the equation $(\sqrt{4489} - \sqrt{2601}) = x^2$,we have $16 = x^2$.Taking the square root of both sides,$x = \sqrt{16} = 4$.Therefore,the value is $4$.

$(\sqrt{4489}-\sqrt{2601})=(?)^{2}$

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