Exercise 5 Page 644

We will find and check the solutions of the given equation.

Finding the Solutions

To solve equations with a variable expression inside a radical, we first want to make sure the radical is isolated. Then we can raise both sides of the equation to a power equal to the index of the radical. Let's try to solve our equation using this method! We now have a quadratic equation, and we need to find its roots. To do it, let's identify the values of $a,$ $b,$ and $c.$

$$\begin{array}{c}{x}^2-13x+30=0\\ \Updownarrow \\ 1x^2+(\text{-}13)x+30=0\end{array}$$

We can see that $a=1,$ $b=\text{-}13,$ and $c=30.$ Let's substitute these values into the Quadratic Formula.

$x=\frac{\text{-}b\pm \sqrt{b^2-4ac}}{2a}$

SubstituteValues

Substitute values

$x=\frac{\text{-}(\text{-}13)\pm \sqrt{(\text{-}13{)}^2-4(1)(30)}}{2(1)}$

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Solve for $x$ and Simplify

NegNeg

$\text{-}(\text{-}a)=a$

$x=\frac{13\pm \sqrt{(\text{-}13{)}^2-4(1)(30)}}{2(1)}$

NegBaseToPosBase

$(\text{-}a{)}^2=a^2$

$x=\frac{13\pm \sqrt{169-4(1)(30)}}{2(1)}$

Multiply

Multiply

$x=\frac{13\pm \sqrt{169-120}}{2}$

SubTerm

Subtract term

$x=\frac{13\pm \sqrt{49}}{2}$

CalcRoot

Calculate root

$x=\frac{13\pm 7}{2}$

Using the Quadratic Formula, we found that the solutions of the given equation are $x=\frac{13\pm 7}{2}.$

$x=\frac{13\pm 7}{2}$
$x_1=\frac{13+7}{2}$	$x_2=\frac{13-7}{2}$
$x_1=\frac{20}{2}$	$x_2=\frac{6}{2}$
$x_1=10$	$x_2=3$

Therefore, the solutions are $x_1=10$ and $x_2=3.$ Let's check them to see if we have any extraneous solutions.

Checking the Solutions

We will check $x_1=10$ and $x_2=3$ one at a time.

$x_1=10$

Let's substitute $x=10$ into the original equation.

$\sqrt{3x-5}=x-5$

Substitute

$x=10$

$\sqrt{3(10)-5}\stackrel{?}{=}10-5$

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Simplify

Multiply

Multiply

$\sqrt{30-5}\stackrel{?}{=}10-5$

SubTerms

Subtract terms

$\sqrt{25}\stackrel{?}{=}5$

CalcRoot

Calculate root

$5=5✓$

We got a true statement, so $x=10$ is a solution.

$x_2=3$

Now, let's substitute $x=3.$

$\sqrt{3x-5}=x-5$

Substitute

$x=3$

$\sqrt{3(3)-5}\stackrel{?}{=}3-5$

Multiply

Multiply

$\sqrt{9-5}\stackrel{?}{=}3-5$

SubTerms

Subtract terms

$\sqrt{4}\stackrel{?}{=}\text{-}2$

CalcRoot

Calculate root

$2\ne \text{-}2\times$

In this case we got a false statement. Therefore, $x=10$ is the only solution to the original equation.