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132 Unit 1: Polynomial, Rational, and Radical Relationships
UNDERSTAND In a radical equation, there is a variable in the radicand . The radicand is the expression inside the radical symbol ( √
__ ), and the index is the root being taken . In the
equation y 5 √__
x , the radicand is x . The equation takes the square root of x, so the index is 2 . Cube roots are also radicals (with an index of 3) .
UNDERSTAND A radical expression is not a polynomial, but polynomials can be used to solve radical equations . You know that linear equations are solved using inverse operations . Some quadratic equations can be solved by taking the square root of both sides . This is because raising to the nth power and taking the nth root are inverse operations . So, to solve a radical equation, raise both sides to the same power .
Consider the equation √__
x 5 3 . The index is 2, so raise both sides to the second power .
√__
x 5 3
( √__
x )2 5 32
x 5 9
Since √__
9 5 3, the answer checks .
Now consider the equation √__
x 5 23 . Solve by squaring both sides .
√__
x 5 23
( √__
x )2 5 (23)2
x 5 9
Since √__
9 23, the answer does not check . It is extraneous .
It is important to check all answers to radical equations since extraneous solutions can occur . You should also be sure to look at the original equation and use number sense . Without solving the equation √
__ x 5 23, you could have concluded that it has no real solution, since a
UNDERSTAND Solving a radical inequality is much like solving a radical equation, but you will have to take some extra steps . As with other inequalities, the solution will usually be a range, which can be graphed on a number line .
Consider the radical inequality √_____
x 1 3 # 2 . Replace the inequality sign with an equals sign, and solve the equation .
√_____
x 1 3 5 2
( √_____
x 1 3 ) 2 5 22 Square both sides .
x 1 3 5 4 Subtract 3 from both sides .
x 5 1
Now, change the equals sign back to the inequality symbol: x # 1 . Graph the solution .
–8 0 8
Remember that you cannot take the square root of a negative number in the set of real
numbers . So, finding the domain of a radical inequality should be your first step in solving
the inequality . With √_____
x 1 3 # 2, the solution only makes sense when the radicand, x 1 3, is
nonnegative . This means x 1 3 $ 0 . Solve this inequality for x .
x 1 3 $ 0 Subtract 3 from both sides .
x $ 23
Graph this solution on a number line .
–8 0 8
The solution to the original inequality is the range where the two solutions overlap .
Check for extraneous solutions . Use the equation .
Try x 5 3 .
√_____
3 1 6 0 3
√__
9 5 3
Try x 5 22 .
√_______
22 1 6 0 22
√__
4 22
The only solution is x 5 3 .
Since the original inequality has the variable on both sides, test values on either side of x to determine which inequality symbol to use (x . 3 or x , 3) .
The solution is x , 3 .
3
1
Replace the inequality sign with an equals sign, and solve the equation .
( √_____
x 1 6 ) 2 5 x2
x 1 6 5 x2
0 5 x22x26
0 5 (x23)(x 1 2)
x 5 3 and x 5 22
2
Graph the complete solution .
Graph the areas of overlap for x , 3 and x $ 26 .
▸ –8 0 8
4
Why is 26 included in the solution to √
_____ x 1 6 . x, but 3 is excluded?
DISCUSS
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136 Unit 1: Polynomial, Rational, and Radical Relationships
The period of a pendulum is the amount of time required for it to swing from one side to the other and back . A pendulum’s period in seconds, P, is related to the length of the pendulum in meters, L, by the following equation:
P 5 2p √___
L ___ 9 .8
A pendulum in a grandfather clock has a period of 2 .5 seconds . Find the length of the pendulum to the nearest tenth of a meter .
PLAN
Substitute the value of the period for P, and solve for L .
SOLVE
Substitute for P into the equation .
5 2p √___
L ___ 9 .8
Solve for L:
( )2 5 ( 2p √___
L ___ 9 .8 ) 2
5 22p
2
L m
CHECK
Substitute the value of L into the equation that relates P and L, and solve .
When L 5 , P .
▸ To the nearest tenth of a meter, the pendulum is m long .
M_569NASE_ALGII_PDF.indd 139 21/07/15 12:55 pm
Practice
140 Unit 1: Polynomial, Rational, and Radical Relationships
22. You can estimate s, the speed in miles per hour, at which a car is moving when it goes into a skid . Use the formula s 5 √
____ 21d , where d is the length of the skid marks in feet .
During an accident, a driver claims to have been driving 42 miles per hour . If the driver’s estimate of his speed is accurate, about how long would the skid marks be?
The skid marks would be about ft long .
The actual skid marks are about 115 ft . Is the driver’s estimate of his speed accurate? Explain .
Solve.
23. EXTEND Solve: √_____
x 1 8 2√__
x 5 2
Isolate the first radical .
√_____
x 1 8 5
Square both sides of the equation .
x 1 8 5
Isolate the radical expression in the resulting equation .
5 √__
x
Square both sides of the resulting equation .
5 x
Show that your answer is correct by substituting it into the original equation .