The solution to the equation sqrt{x + 10} + s | vedclass

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(B) Given equation: $\sqrt{x + 10} + \sqrt{x - 2} = 6$ Let $u = \sqrt{x + 10}$ and $v = \sqrt{x - 2}$. Then $u^2 = x + 10$ and $v^2 = x - 2$. Subtract

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(B) Given equation: $\sqrt{x + 10} + \sqrt{x - 2} = 6$ Let $u = \sqrt{x + 10}$ and $v = \sqrt{x - 2}$. Then $u^2 = x + 10$ and $v^2 = x - 2$. Subtracting the two: $u^2 - v^2 = (x + 10) - (x - 2) = 12$. We know $(u - v)(u + v) = 12$. Since $u + v = 6$,we have $(u - v)(6) = 12$,which implies $u - v = 2$. Now we have a system: $u + v = 6$ $u - v = 2$ Adding these: $2u = 8 \implies u = 4$. Since $u = \sqrt{x + 10} = 4$,squaring both sides gives $x + 10 = 16$,so $x = 6$. Checking the value: $\sqrt{6 + 10} + \sqrt{6 - 2} = \sqrt{16} + \sqrt{4} = 4 + 2 = 6$. The solution is $x = 6$.

The solution to the equation $\sqrt{x + 10} + \sqrt{x - 2} = 6$ is:

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