How do you solve $4 r^{2} + 1 = 325$ $4r^2+1=325$ ?

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Answer 1 · 2017-01-20T16:56:37

$r^{2} = 81$ $r^2 = 81$ which means $r$ $r$ is either 9 or -9.

Given the equation $4 r^{2} + 1 = 325$ $4r^2+1=325$ , we are going to look for $r^{2}$ $r^2$

So to find $r^{2}$ $r^2$ we are going to isolate it on one side of the equation.

Step 1:

Subtract 1 from both sides of the equation

$4 r^{2} + 1 - 1 = 325 - 1$ $4r^2+1 -1 = 325-1$

So we get $4 r^{2} = 324$ $4r^2 = 324$

Step 2:

Divide both sides by 4

$4 \frac{r^{2}}{4} = \frac{324}{4}$ $4r^2/4 = 324/4$

So now we isolated have $r^{2}$ $r^2$ isolated

$r^{2} = 81$ $r^2 = 81$

$r = \pm 9$ $r=+-9$

How do you solve 4r^2+1=3254r2+1=3254r^2+1=325?