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Back to the class

__Quiz 25__

**1.)** Rationalize the denominator of
$$\dfrac{1}{\sqrt{5}}.$$

*Solution:* Calculate
$$\dfrac{1}{\sqrt{5}} = \dfrac{1}{\sqrt{5}} \dfrac{\sqrt{5}}{\sqrt{5}} = \dfrac{\sqrt{5}}{5}.$$

**2.)** Rationalize the denominator of
$$\dfrac{2}{5+\sqrt{2}}.$$
*Solution:* Calculate
$$\dfrac{2}{5+\sqrt{2}} = \dfrac{2}{5+\sqrt{2}} \dfrac{5-\sqrt{2}}{5-\sqrt{2}} = \dfrac{10-2\sqrt{2}}{25-2} = \dfrac{10-2\sqrt{2}}{23}.$$
**3.)** Solve the equation
$$5 \sqrt{y} = -2.$$
*Solution:* First isolate $\sqrt{y}$ by dividing by $5$:
$$\sqrt{y} = -\dfrac{2}{5}.$$
Now square both sides to get
$$y = \dfrac{4}{25}.$$
Is this the solution? (RECALL: we must *always* check the solution of a radical equation back in the original). Plug this value into the original equation to get
$$5 \sqrt{\dfrac{4}{25}} \stackrel{?}{=} -2.$$
$$5 \dfrac{2}{5} \stackrel{?}{=} -2.$$
$$2 \stackrel{?}{=} -2.$$
No!! This means that the equation has **no solution**.

**4.)** Solve the equation
$$\sqrt{7t-9} = \sqrt{t+3}.$$
*Solution:* Square both sides to get
$$7t-9 = t+3.$$
Subtract $t$ and add $9$ to get
$$6t = 12.$$
Divide by $6$ to get
$$t = 2.$$
**CHECK** this solution by plugging it into the original:
$$\sqrt{7(2)-9} \stackrel{?}{=} \sqrt{2+3}$$
and simplify
$$\sqrt{14-9} \stackrel{?}{=} \sqrt{5}$$
$$\sqrt{5} \stackrel{?}{=} \sqrt{5}.$$
Yes! It's true. Therefore the solution is $t=2$.

**5.)** Solve the equation
$$1+2\sqrt{y-1} = y.$$
*Solution:* First we need to isolate $\sqrt{y-1}$. To do this, subtract $1$ and then divide by $2$ to get
$$\sqrt{y-1} = \dfrac{y-1}{2}.$$
Now square both sides to get
$$y-1 = \dfrac{(y-1)^2}{4}.$$
Expand the binomial on the right-hand side to get
$$y-1 = \dfrac{y^2-2y+1}{4}.$$
It is easier to go ahead and multiply by $4$ to get
$$4y-4 = y^2-2y+1.$$
Now get all terms on the same side by subtracting $4y$ and adding $4$:
$$0 = y^2-6y+5.$$
This is a quadratic equation that we can solve by factoring:
$$0 = (y-5)(y-1),$$
and so the solution is $y=5$ or $y=1$. Now we must check both of these solutionis in the original equation:

__Check $y=1$__

$$1+2\sqrt{1-1}\stackrel{?}{=}1$$
yes! It works.

__Check $y=5$__

$$1 + 2 \sqrt{5-1} \stackrel{?}{=} 5.$$
Yes! It works! (the square root on the left becomes $\sqrt{4}$ which is $2$ and $1+2\cdot 2 = 5$).

Therefore we have the two solutions $y=5$ and $y=1$.