Question

if the solutions to $4x^{2}+1 = 81$ are $pm gsqrt{5}$, what is the value of $g$?
$g = \square$
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Explanation:

Step1: Solve the equation for \(x^2\)

Start with the equation \(4x^2 + 1 = 81\). Subtract 1 from both sides:
\(4x^2 = 81 - 1\)
\(4x^2 = 80\)
Then divide both sides by 4:
\(x^2=\frac{80}{4}\)
\(x^2 = 20\)

Step2: Find the square root of \(x^2\)

Take the square root of both sides. We know that if \(x^2 = 20\), then \(x=\pm\sqrt{20}\). Simplify \(\sqrt{20}\):
\(\sqrt{20}=\sqrt{4\times5}=\sqrt{4}\times\sqrt{5} = 2\sqrt{5}\)
The solutions are given as \(\pm g\sqrt{5}\), so comparing \(\pm 2\sqrt{5}\) with \(\pm g\sqrt{5}\), we can see that \(g = 2\).

Answer:

\(2\)