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$Solve the following equation for y. {:[sqrt(y^(2)+4)=sqrt(4y)],[y=]:}$

Solve the following equation for $y$ . $\newline$ $\begin{array}{l} \sqrt{y^{2}+4}=\sqrt{4 y} \\ y= \end{array}$

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Domain and range of square root functions: equations

Full solution

Q. Solve the following equation for

y

.

\newline

\begin{array}{l} \sqrt{y^{2}+4}=\sqrt{4 y} \\ y= \end{array}

Write Equation: Write down the given equation. $\newline$ We have the equation $\sqrt{y^2 + 4} = \sqrt{4y}$ .
Square Both Sides: Square both sides of the equation to eliminate the square roots. $\newline$ $(\sqrt{y^2 + 4})^2 = (\sqrt{4y})^2$ $\newline$ This gives us $y^2 + 4 = 4y$ .
Rearrange and Solve: Rearrange the equation to set it to zero and find the values of $y$ . $y^2 - 4y + 4 = 0$
Factor Quadratic Equation: Factor the quadratic equation. $\newline$ $(y - 2)^2 = 0$
Solve for $y$ : Solve for $y$ by taking the square root of both sides. $\newline$ $y - 2 = 0$ $\newline$ So, $y = 2$ .
Check Solution: Check the solution in the original equation to ensure it does not produce any math errors. $\newline$ $\sqrt{2^2 + 4} = \sqrt{4\times 2}$ $\newline$ $\sqrt{4 + 4} = \sqrt{8}$ $\newline$ $\sqrt{8} = \sqrt{8}$ $\newline$ This is true, so $y = 2$ is a valid solution.

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