PreCalculus 11 – Week 5

This week in Precalculus 11 we learned how to solve radical equations. Radical equations look like $2\sqrt {x}$ = 14.

The first thing you have to do to solve radical equations is set restrictions, to do that you cant have a coefficient in front of the root. In the case of $2\sqrt {x}$ = 14, we need to divide by the coefficient so $2\sqrt {x}$ = 14 becomes $\frac{2\sqrt{x}}{2} = \frac{14}{2}$ . So now that we have $\sqrt {x}$ = 7 we can set restrictions. Restrictions are a set of rules of what x can equal. In this equations, the restrictions would look like this, x>0. X has to be greater than zero because this is a square root which is always positive. Now that we have the restrictions and $\sqrt {x}$ = 7, we can now solve this equation. To solve this equation we must isolate x, to do this we must square both sides of the equation.

$(\sqrt {x})^2 = (7)^2$ .

X = 49

If the equation looks like this $\sqrt {2x+7}$ = 5 we can set restrictions right away, for this equation the restriction would be $\frac{2x}{2} >\frac{7}{2}$ = x> $\frac{7}{2}$ . Now that we have the restrictions we can solve this equation.

$\sqrt {2x+7}$ = 5

$(\sqrt {2x+7})^2 = (5)^2$

2x+7 = 25

2x = 18

$\frac{2x}{2} = \frac{18}{2}$

X = 2

PreCalculus 11 – Week 5

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