Solve If sqrt(2025) + sqrt(0.0612 + x) = 45.25 , find the value of x | Equatix - Math Solver

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a⋅(b-2)=3b

3(r+2s)=2t-4

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Question

$$\sqrt { 2025 } + \sqrt { 0.0612 + x } = 45.25 , x$$

Solve for x

$x=0.0013$

Steps for Solving Radical Equations

Subtract $\sqrt{2025}$ from both sides of the equation.

$$\sqrt{x+0.0612}+\sqrt{2025}-\sqrt{2025}=45.25-\sqrt{2025}$$

Subtracting $\sqrt{2025}$ from itself leaves $0$.

$$\sqrt{x+0.0612}=45.25-\sqrt{2025}$$

Subtract $\sqrt{2025}$ from $45.25$.

$$\sqrt{x+0.0612}=0.25$$

Square both sides of the equation.

$$x+0.0612=0.0625$$

Subtract $0.0612$ from both sides of the equation.

$$x+0.0612-0.0612=0.0625-0.0612$$

Subtracting $0.0612$ from itself leaves $0$.

$$x=0.0625-0.0612$$

Subtract $0.0612$ from $0.0625$ by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$x=0.0013$$