ORCID iD icon

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 $y=1$.