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Unformatted text preview: e the xcoordinate from
the points of intersection. SOLVING RADICAL EQUATIONS ALGEBRAICALLY
Radical equations will be easier to solve without the radical. x+5 = 2 9− o Isolate the square root. − example: Solve 9− x+5 = 2
x + 5 = −7
x+5 = 7 ( o Square both sides of the equation.
o Simplify.
o Solve for x .
o Check this root.
Answer: exercise: Solve 9− 2
x + 5 ) = (7)
x + 5 = 49
x = 44 2 (44) + 5 = 2 x = 44 19 − 2 1 − 3 x = 11 [Answer: −5] Unit 4: Day 4 notes  Radical Equations (Part 1) Page 2 of 2 x − x2 + x − 1 = 0 exercise: Solve [Answer: 1 o r 1]
2 EXTRANEOUS ROOTS
An extraneous root is a value the algebraic steps may produce as roots; however it does
not satisfy the original equation and so cannot be a root.
example: Solve x+5 − x = 3 x+5 − x = 3
x+5 = x + 3 o Isolate the square root.
o Square both sides of the equation. ( o Simplify. 2
x + 5 ) = (x + 3)
x + 5 = x 2 + 6x + 9 o Solve for x . 0
0
x+4 = 0
x = −4 Check the roots.
Answer: −4 + 5 − (−4) ≠ 3 or
or = x 2 + 5x + 4
= (x + 4)(x + 1)
(x + 1) = 0
x = −1 −1 + 5 − (−1) = 3 x = −1
5 − 2 x = −7 [Answer: −2] 1 − x2 − x + 1 = 0 [Answer: 1] exercise: Solve 2x − exercise: Solve 2 PRECALCULUS 11 Unit 4 – Day 5: RADICAL EQUATIONS (Part 2) RADICAL EQUATIONS WITH TWO RADICAL TERMS
x −1 − 5 − x = 0 is an example of a radical equation with two radical terms. With two radical terms, there are two restrictions to consider: x − 1 ≥ 0 and 5 − x ≥ 0
which is x ≥ 1 and x ≤ 5 , ∴ 1 ≤ x ≤ 5 . Radical equations will be easier to solve without the radicals; isolate a radical and
square both sides of the equation. example: Solve x −1 − 5− x = 0 x −1 − x −1 = o Isolate the first square root.
o Square both sides of the equation. ( x −1 ) =
2 x−1
2x
x o Simplify.
o Solve for x .
(3) − 1 − o Check this root.
Answer: 5− x = 0
5− x ( 5− x )2 = 5−x
=6
=3 5 − (3) = 0 x=3 exercise: Solve x+6 − 3 − 2x = 0 [Answer: −1] exercise: Solve x +1 − x+7 = 0 [Answer: φ] Unit 4: Day 5 notes  Radical Equations (Part 2) Page 2 of 2 Isolating one of the radicals and squaring both sides of the equation may not eliminate
both radical symbols. If that happens work on one radical at a time. 4x + 5 − example: Solve 2x − 1 = 2 4x + 5 − 2x − 1 = 2 4x + 5 = 2 + o Isolate the first square root. ( o Square both sides. 2x − 1 4 x + 5 ) = (2 + 2 x − 1 )
2 2 4x + 5 o Isolate the second square root. = 4 + 4 2 x − 1 + 2x − 1 4x + 5 o Simplify. = 2x + 3 + 4 2 x − 1 2x + 2 = 4 2 x − 1 o Simplify x + 1 = 2 2x − 1 o Square both sides. (x + 1)2 = ( 2 2 x − 1 ) 2 x2 + 2x + 1 = 4(2x − 1)
x 2 + 2x + 1 = 8 x − 4
x2 − 6x + 5 = 0 o Simplify (x − 1)(x − 5) = 0
x−1 = 0
or
x−5 = 0
x=1
or
x=5 o Solve for x . 4(1) + 5 − Answer: x=1 or exercise: Solve x−5 + exercise: Solve x−5 +3 = 2(1) − 1 = 2 4(5) + 5 − Check the roots. 2(5) − 1 = 2 x=5 x+4 = 9 2x + 7 [Answer: 21] [Answer: 9 or 21]...
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 Fall '11
 Aytona
 Calculus, PreCalculus, Radicals

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