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Solve for xX - 4 =(2x - 5Step 1 of 5Using the squaring property of equality, eliminate the radical from the equation.Square both sides of the equation to eliminate the square root. Then, expand the square of the binomial usingthe formula, (x - y)2 = x2 - 2xy + y?(x - 4)2 =(v2x - 5)(x - 4)2 = 2x -x² - 2(x)1) + 42 = 2x -X + 16 = 2x -

1 Answer

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Step 1

Expand both sides of the equation.


\begin{gathered} (x-4)^2=(\sqrt[]{2x-5)}^2 \\ (x-4)\text{ ( x-4) = (}\sqrt[]{2x-5)}\text{ (}\sqrt[]{2x-5)} \end{gathered}


\begin{gathered} x^2-4x-4x\text{ +16 = }(2x-5)^{(1)/(2)}(2x-5)^{(1)/(2)} \\ x^2-8x+16=(2x-5)^{(1)/(2)+(1)/(2)} \end{gathered}


\begin{gathered} x^2-8x+16=(2x-5)^1 \\ x^2-8x\text{ +16 = 2x-5} \end{gathered}

Step 2

Apply the given formula to the answer in step 1


\begin{gathered} y\text{ }=\text{ 4} \\ x^2_{}-2(x)(4)+4^2=2x-5 \end{gathered}

Hence the required answer is


x^2-2(x)(4)+4^2=2x-5

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