find the value of y when $x = 3$ if $y^{2} = x + 1$

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Solution

The correct option is D 2
$y^{2} = x + 1 y^{2} = 3 + 1 y^{2} = 4 y = \sqrt{4} y = 2$

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Similar questions

y = (x + \sqrt{x^{2} - y^{2}}

y = 3

x^{2} + y^{2} = 25

x = y^{2}

y = \frac{5}{k}

x

k = \frac{1}{2}

x = \sqrt{3} + \sqrt{2}

y = \frac{1}{\sqrt{3} + \sqrt{2}}

(x + y)^{2} + (x - y)^{2}

x = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}

y = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}

x^{2} + y^{2}

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