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Correct to the nearest tenth, the solution to the radical equation
2sqrt(2x-1)-8=0 is

Correct to the nearest tenth, the solution to the radical equation $2 \sqrt{2 x-1}-8=0$ is

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Relationship between squares and square roots

Full solution

Q. Correct to the nearest tenth, the solution to the radical equation

2 \sqrt{2 x-1}-8=0

is

Isolate square root term: First, we need to isolate the square root term on one side of the equation. $\newline$ Starting with the equation: $2\sqrt{2x-1} - 8 = 0$ $\newline$ Add $8$ to both sides to isolate the square root term: $2\sqrt{2x-1} = 8$
Divide to solve: Next, we divide both sides of the equation by $2$ to solve for the square root term. $\newline$ $\frac{2\sqrt{2x-1}}{2} = \frac{8}{2}$ $\newline$ This simplifies to: $\sqrt{2x-1} = 4$
Square both sides: Now, we square both sides of the equation to eliminate the square root. $\newline$ $(\sqrt{2x-1})^2 = 4^2$ $\newline$ This results in: $2x - 1 = 16$
Add to isolate x: We then add $1$ to both sides of the equation to isolate the term with $x$ . $\newline$ $2x - 1 + 1 = 16 + 1$ $\newline$ This simplifies to: $2x = 17$
Divide to solve for $x$ : Finally, we divide both sides by $2$ to solve for $x$ . $\frac{2x}{2} = \frac{17}{2}$ This gives us: $x = 8.5$

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