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Solving Radical Expressions

by Camron
(Decatur, Alabama)










































@RT{x+4}+3=@RT{2x+25}

Solve the equation containing radicals for x.

As a general observation, to eliminate the square root signs on each side of the equation will require you to square each side of the equation.

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Nov 18, 2011
Solving Radical Equations
by: Staff

Question:
by Camron
(Decatur, Alabama)

@RT{x+4}+3=@RT{2x+25}


Answer:

@RT{x+4}+3=@RT{2x+25}

I assume the @RT stands for a square root sign so that:

√{x + 4} + 3 = √{2x + 25}

Square each side of the equation

[√{x + 4} + 3]² = √[{2x + 25}]²

x+6sqrt(x+4) + 13 = 2x + 25


Subtract x from each side of the equation

x - x + 6sqrt(x+4) + 13 = 2x - x + 25

0 + 6 sqrt(x + 4) + 13 = x + 25


Subtract 13 from each side of the equation

6sqrt(x + 4) + 13 = x + 25

6sqrt(x + 4) + 13 - 13 = x + 25 - 13

6sqrt(x + 4) + 0 = x + 12

6sqrt(x + 4) = x + 12


Square each side of the equation

[6sqrt(x + 4)]² = (x + 12)²


36(x + 4) = (x + 12)²

36(x + 4) = x² + 24x + 144


Subtract 144 from each side of the equation

36x + 144 - 144 = x² + 24x + 144 - 144

36x + 0 = x² + 24x + 0

36x = x² + 24x

x² + 24x = 36x


Subtract 36x from each side of the equation

x² + 24x - 36x = 36x - 36x

x² + 24x - 36x = 0

x² - 12x = 0


Factor the left hand side of the equation

x(x - 12) = 0


Since x(x - 12) = 0, either x = 0, or x - 12 = 0, or both x and x - 12 must = 0


Solve for the value of x which causes each factor to = 0


1st factor

x = 0


2nd factor

x - 12 = 0

x - 12 + 12 = 0 + 12

x + 0 = 0 + 12

x = 12



the final answer is:

x ∈ {0, 12}


Check your work:


For x = 0

Substitute 0 for x in the original equation


√{x + 4} + 3 = √{2x + 25}

√{0 + 4} + 3 = √{2*0 + 25}

√{4} + 3 = √{0 + 25}

√{4} + 3 = √{25}

√{4} + 3 = √{25}

2 + 3 = 5

5 = 5, YES → x = 0 is a VALID SOLUTION



For x = 12

Substitute 12 for x in the original equation


√{x + 4} + 3 = √{2x + 25}

√{12 + 4} + 3 = √{2*12 + 25}

√{16} + 3 = √{24 + 25}

√{16} + 3 = √{49}

4 + 3 = 7

7 = 7, YES → x = 12 is a VALID SOLUTION










Thanks for writing.

Staff
www.solving-math-problems.com



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