Big Ideas Math Algebra 1, 2015
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Big Ideas Math Algebra 1, 2015 View details
3. Solving Radical Equations
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Exercise 1 Page 564

Think about extraneous solution.

See solution.

Practice makes perfect
We want to explain why we should check every solution of a radical equation. To do so, we will write an example radical equation and examine its solutions. x = sqrt(3-2x)When we solve this equation, we find two different solutions: x= 1 and x= -3. The solving of this can be seen at the bottom of the solution. Now, we will check the solutions by substituting them back into the equation. Let's begin with x= 1.
x = sqrt(3-2x)
1 ? = sqrt(3-2( 1))
1 ? = sqrt(3-2)
1? = sqrt(1)
1 = 1 âś“
Since we found an identity, x= 1 is a real solution for original equation. Next, we will check for x= -3.
x = sqrt(3-2x)
-3 ? = sqrt(3-2( -3))
-3 ? = sqrt(3-(-6))
-3 ? = sqrt(9)
-3 ≠ 3 *
Notice that x= -3 does not correctly solve our equation, even though it was found as a solution. This is called an extraneous solution. It is very common that raising the sides of a radical equation to an exponent can introduce extraneous solutions. With this in mind, it is good practice to check every solution of a radical equation.

Showing Our Work

Solution to the Example Equation
To solve the example equation, we need to first eliminate the radical. To do so, we will square both sides of the equation and then solve it for x.
x =sqrt(3-2x)
x ^2=(sqrt(3-2x))^2
x^2=3-2x
â–Ľ
Rewrite
x^2+2x=3
x^2+2x-3=0
â–Ľ
Factor
x^2+3x-x-3=0
x(x+3)-x-3=0
x(x+3)-1(x+3)=0
(x+3)(x-1)=0
The equation is now in factored form. Therefore, we will use the Zero Product Property to solve it. (x+3)(x-1)=0 ⇓ x= 1 or x= -3