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The main idea behind solving equations containing square roots is to raise to power 2 in order to clear the square root using the property
( sqrt(x) ) 2 = x.
The above holds only for x greater than or equal to 0. In solving equations we square both sides of the equations and instead of putting similar conditions we check the solutions obtained.
Example 1 : Find all real solutions to the equation
sqrt ( x + 1) = 4
Solution to Example 1:
- Given
sqrt ( x + 1) = 4
- We raise both sides to power 2 in order to clear the square root.
[ sqrt ( x + 1) ] 2 = 4 2
- and simplify
x + 1 = 16
- Solve for x.
x = 15
- NOTE: Since we squared both sides without putting any conditions, extraneous solutions may be introduced, checking the solutions is necessary.
Left side (LS) of the given equation when x = 15
LS = sqrt (x + 1) = sqrt (15 + 1) = 4
Right Side (RS) of the given equation when x = 15
RS = 4
- For x = 15, the left and the rigth sides of the given equation are equal: x = 15 is a solution to the given equation.
Example 2 : Find all real solutions to the equation
sqrt ( 3 x + 1) = x - 3
Solution to Example 2:
- Given
sqrt ( 3 x + 1) = x - 3
- We raise both sides to power 2.
[ sqrt ( 3 x + 1) ] 2 = (x - 3) 2
- and simplify.
3 x + 1 = x 2 - 6 x + 9
- Write the equation with right side equation to 0.
x 2 - 9 x + 8 = 0
- It is a quadratic equation with 2 solutions
x = 8 and x = 1
- NOTE: Since we squared both sides , extraneous solutions may be introduced, checking the solutions in the original equation is necessary.
1. check equation for x = 8.
Left side (LS) of the given equation when x = 8
LS = sqrt ( 3 x + 1) = sqrt (3 * 8 + 1) = 5
Right Side (RS) of the given equation when x = 8
RS = x - 3 = 8 - 3 = 5
2. check equation for x = 1.
Left side (LS) of the given equation when x = 8
LS = sqrt ( 3 x + 1) = sqrt (3 * 1 + 1) = 2
Right Side (RS) of the given equation when x = 8
RS = x - 3 = 1 - 3 = -2
- For x = 8 the left and right sides of the equation are equal and x = 8 is a solution to the given equation. x = 1 is not a solution to the given equation; it is an extraneous solution introduced because of the raising to power 2.
Exercises:(answers further down the page)
Solve the following equations
1. sqrt(2 x + 15) = 5
2. sqrt(4 x - 3) = x - 2
Solutions to above exercises
1. x = 5
2. x = 7
More references and links on how to Solve Equations, Systems of Equations and Inequalities.
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