SOLUTION: I cannot solve the following, please help and show step by step if possible, thank you {{{matrix(1,3,sqrt(7x+29), "=", x+3)}}}

Algebra ->  Radicals -> SOLUTION: I cannot solve the following, please help and show step by step if possible, thank you {{{matrix(1,3,sqrt(7x+29), "=", x+3)}}}      Log On


   

Question 176651This question is from textbook Elementary and Intermediate Algebra
: I cannot solve the following, please help and show step by step if possible, thank you
matrix%281%2C3%2Csqrt%287x%2B29%29%2C+%22=%22%2C+x%2B3%29 This question is from textbook Elementary and Intermediate Algebra

Found 2 solutions by jim_thompson5910, Edwin McCravy:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt%287x%2B29%29=x%2B3 Start with the given equation.


7x%2B29=%28x%2B3%29%5E2 Square both sides


7x%2B29=x%5E2%2B6x%2B9 FOIL


0=x%5E2%2B6x%2B9-7x-29 Subtract 7x from both sides. Subtract 29 from both sides.


0=x%5E2-x-20 Combine like terms.


0=%28x-5%29%28x%2B4%29 Factor the right side



Now set each factor equal to zero:

x-5=0 or x%2B4=0


x=5 or x=-4 Now solve for x in each case


So the possible answers are x=5 or x=-4


However, if you plug in x=-4, you'll find that the original equation will not be true. So x=-4 is NOT a solution


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Answer:

So the only solution is x=5

Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
I cannot solve the following, please help and show step by step if possible, thank you
matrix%281%2C3%2Csqrt%287x%2B29%29%2C+%22=%22%2C+x%2B3%29 

Since the radical is already isolated on the left,
square both sides.

matrix%281%2C3%2C%28sqrt%287x%2B29%29%29%5E2%2C+%22=%22%2C+%28x%2B3%29%5E2%29

matrix%281%2C3%2Csqrt%287x%2B29%29sqrt%287x%2B29%29%2C+%22=%22%2C+%28x%2B3%29%28x%2B3%29%29

matrix%281%2C3%2C7x%2B29%2C+%22=%22%2C+x%5E2%2B3x%2B3x%2B9%29

matrix%281%2C3%2C7x%2B29%2C+%22=%22%2C+x%5E2%2B6x%2B9%29

Get 0 on the left by subtracting 7x%2B29 from
both sides:

+matrix%281%2C3%2C0%2C+%22=%22%2C+x%5E2-x-20%29

or

matrix%281%2C3%2Cx%5E2-x-20%2C+%22=%22%2C0%29

Factor left side:

matrix%281%2C3%2C%28x-5%29%28x%2B4%29%2C+%22=%22%2C0%29

Use the principle of zero factors:

  

But we have to check each of those,
because squaring both sides, or multiplying
both sides by a variable, may result in
bogus solutiona, called "extraneous"
solutions.

Checking x=5

matrix%281%2C3%2Csqrt%287x%2B29%29%2C+%22=%22%2C+x%2B3%29
matrix%281%2C3%2Csqrt%287%285%29%2B29%29%2C+%22=%22%2C+%285%29%2B3%29
matrix%281%2C3%2Csqrt%2835%2B29%29%2C+%22=%22%2C+5%2B3%29
matrix%281%2C3%2Csqrt%2864%29%2C+%22=%22%2C+8%29
matrix%281%2C3%2C8%2C+%22=%22%2C+8%29

That checks, so x=5 is a solution.

Checking x=-4

matrix%281%2C3%2Csqrt%287x%2B29%29%2C+%22=%22%2C+x%2B3%29
matrix%281%2C3%2Csqrt%287%28-4%29%2B29%29%2C+%22=%22%2C+%28-4%29%2B3%29
matrix%281%2C3%2Csqrt%28-28%2B29%29%2C+%22=%22%2C+-4%2B3%29
matrix%281%2C3%2Csqrt%281%29%2C+%22=%22%2C+-1%29
matrix%281%2C3%2C1%2C+%22=%22%2C+-1%29

That does not check, so x=4 is not 
a solution, and is called "extraneous".

The only genuine solution is x=5

Edwin