McGraw Hill Glencoe Algebra 1, 2012
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McGraw Hill Glencoe Algebra 1, 2012 View details
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Exercise 12 Page 667

Raise both sides of the equation to a power equal to the index of the radical.

x=3

Practice makes perfect
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!
sqrt(4x-3)=6-x
(sqrt(4x-3))^2=(6-x)^2
4x-3=(6-x)^2
4x-3 = 36-12x+x^2
LHS-(4x-3)=RHS-(4x-3)
-3 = 36-16x+x^2
0 = 39-16x+x^2
0 = x^2-16x+39
x^2 -16x+39=0
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. x^2 -16x +39 = 0 ⇔ 1x^2+( - 16)x+ 39=0We can see that a= 1, b= - 16, and c= 39. Let's substitute these values into the Quadratic Formula.
x=- b±sqrt(b^2-4ac)/2a
x=- ( -16)±sqrt(( - 16)^2-4( 1)( 39))/2( 1)
Solve for x and Simplify
x=16±sqrt((- 16)^2-4(1)(39))/2(1)
x=16±sqrt(256-4(1)(39))/2(1)
x=16±sqrt(256-156)/2
x=16±sqrt(100)/2
x=16± 10/2
Using the Quadratic Formula, we found that the solutions of the given equation are x= 16± 10 2. Therefore, the solutions are x_1= 13 and x_2=3. Let's check them to see if we have any extraneous solutions.

Checking the Solutions

We will check x_1=13 and x_2=3 one at a time.

x_1=13

Let's substitute x= 13 into the original equation.
sqrt(4x-3)=6-x
sqrt(4( 13)-3)? =6- 13
Simplify
sqrt(52-3)? =6-13
sqrt(49)? = -7
7 ≠ -7 *
We got a false statement, so x=13 is an extraneous solution.

x_2=3

Now, let's substitute x=3.
sqrt(4x-3)=6-x
sqrt(4(3)-3)? =6-3
Simplify
sqrt(12-3)? =6-3
sqrt(9)? = 3
3=3 ✓
In this case we got a true statement. Therefore, x=3 is the only solution of the original equation.