6.6 Solve Radical
Equations
VOCABULARY
Radical
equation
An equation with a radical that
has variables in the radicand
SOLVING RADICAL
EQUATIONS
To solve a radical equation,
follow these steps:
Step 1 _Isolate_ the radical on one side of the equation, if
necessary.
Step 2 Raise each side of the equation
to the same _power_ to eliminate the radical and obtain a linear,
quadratic, or other polynomial equation.
Step 3 _Solve_ the polynomial
equation using techniques you learned in
previous chapters. Check your solution.
Example 1
Solve a radical equation
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Solve
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Write original equation. |
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Square each side to eliminate
the radical. |
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x+
6 = _9_ |
Simplify. |
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_x_ = _3_ |
Subtract
_6_ from each side. |
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The
solution is _3_. Check this in the original equation |
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.
Solve the equation. Check your solution.
-3
Example 2
Solve an
equation with a rational exponent
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(3x + 4)2/3
= 16 |
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[(3x+ 4)2/3]3/2 = 163/2 |
Raise each side to the Power |
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3x + 4 = (161/2)3 |
Apply properties of exponents. |
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3x+ 4 = 64 |
Simplify. |
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3x = 60 |
Subtract
_4_ from each side. |
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_x_ = 20 |
Divide
each side by _3_. |
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The solution is _20_.
Check this in the original equation |
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.
Example 3
Solve an
equation with an extraneous solution
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x -2 = |
Original equation |
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Square each side. |
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x2 - 4x+ 4 = x+ 10 |
Expand left side and simplify
right side. |
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x2 - 5x - 6 = 0 |
Write in standard form. |
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(x- 6)(x
+ 1) = 0 |
Factor. |
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x- 6 = 0 or x+ 1 = 0 |
Zero product property |
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x = _6_ or x = -1 |
Solve for x |
CHECK |
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Check x = _6_. |
Check x = _-1_. |
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__4__=__4__ |
_-3 ¹ 3_ |
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The only
solution is __4__ (The apparent solution __- 1__ is extraneous.)
Example 4
Solve an
equation with two radicals
Solve
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Write original equation. |
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Square each side. |
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Expand left side and simplify right side. |
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Isolate radical expression. |
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Divide each side by 4. |
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Square each side again. |
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x + 6 = x2 |
Simplify. |
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0 = x2
- x - 6 |
Write in standard form. |
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0 = (x
- 3)(x + 2) |
Factor. |
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x - 3 = 0 or x + 2
= 0 |
Zero product
property |
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x = 3 or x = -2 |
Solve for x |
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CHECK Check x = 3 . |
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5 = 1 |
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Check x = - _-2
. |
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_4__=__4_ |
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The only solution is -2_. (The apparent
solution _3_ is extraneous.)