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In order to solve the equation, we must square root both sides. We must not forget that we also get a negative solution, since the product of two negative numbers is positive.

Both $x=-9$ and $x=9$ are solutions to the equation.

b

Before we square root the equation, we need to isolate the $x_{2}$-term on one side.

$2x_{2}−3=47$

AddEqn$LHS+3=RHS+3$

$2x_{2}=50$

DivEqn$LHS/2=RHS/2$

$x_{2}=25$

SqrtEqn$LHS =RHS $

$x=±25 $

SimpRootSimplify root

$x=±5$

We **can** take the square root before adding $3$ or before dividing by $2.$ However, the solution becomes a bit more complicated. Let's square root the equation before dividing by $2.$ We must keep in mind that we must square root **the entire LHS**.

$2x_{2}−3=47$

AddEqn$LHS+3=RHS+3$

$2x_{2}=50$

SqrtEqn$LHS =RHS $

$2x_{2} =±50 $

SqrtProd$a⋅b =a ⋅b $

$2 x=±50 $

DivEqn$LHS/2 =RHS/2 $

$x=±2 50 $

DivSqrt$b a =ba $

$x=±250 $

SimpQuotSimplify quotient

$x=±25 $

SimpRootSimplify root

$x=±5$

We still end up with the same result, even though the process was more complicated.

c

First we will square root both sides.

Both $x=-90 $ and $x=90 $ are solutions to the equation. If we want to, we can simplify the right side further, so that we get a smaller number under the root sign.$x=±9⋅10 $

SplitIntoFactorsSplit into factors

$x=±9⋅10 $

SqrtProd$a⋅b =a ⋅b $

$x=±9 ⋅10 $

SimpRootSimplify root

$x=±310 $