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Adding Radicals.











































I am correcting questions on an Algebra II pre-test, and can't seem to understand how to work the equation √(5+x)+√(x)=5. I know the answer is 4 but I don't know how to get there.

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Aug 22, 2011
Adding Radicals
by: Staff


The question:

I am correcting questions on an Algebra II pre-test, and can't seem to understand how to work the equation √(5+x)+√(x)=5. I know the answer is 4 but I don't know how to get there.


The answer:

√(5+x) + √(x) = 5

As you already know, you can only add radicals if the radicand is the same. That is what makes this problem difficult.

Here is how you solve the problem analytically:

√(5+x) + √(x) = 5

Subtract √(x) from each side of the equation

√(5+x) + √(x) - √(x) = 5 - √(x)

√(5+x) + 0 = 5 - √(x)

√(5+x) = 5 - √(x)

Subtract 5 from each side of the equation

√(5+x) - 5 = 5 - 5 - √(x)

√(5+x) - 5 = 0 - √(x)

√(5+x) - 5 = - √(x)

Square both sides of the equation

[√(5+x) - 5]² = [- √(x)]²

[√(5+x) - 5]² = [-1]² * [√(x)]²

[√(5+x) - 5]² = 1 * [√(x)]²

[√(5+x) - 5]² = [√(x)]²

[√(5+x) - 5]² = x

5 + x - 10√(5+x) + 25 = x

5 + 25 + x - 10√(5+x) = x

30 + x - 10√(5+x) = x

Subtract x from each side of the equation

30 + x - x - 10√(5+x) = x - x

30 + 0 - 10√(5+x) = x – x

30 - 10√(5+x) = x – x

30 - 10√(5+x) = 0

Add 10√(5+x) to each side of the equation

30 - 10√(5+x) + 10√(5+x) = 0 + 10√(5+x)

30 + 0 = 0 + 10√(5+x)

30 = 0 + 10√(5+x)

30 = 10√(5+x)

Divide each side of the equation by 10

30/10 = [10√(5+x)]/10

3 = [10√(5+x)]/10

3 = [√(5+x)]*[10/10]

3 = [√(5+x)]*[1]

3 = √(5+x)

Square each side of the equation

3² = [√(5+x)]²

9 = [√(5+x)]²

9 = 5 + x

Subtract 5 from each side of the equation

9 - 5 = 5 - 5 + x

4 = 5 - 5 + x

4 = 0 + x

4 = x

The final answer is: x = 4


You can also solve this equation graphically.


(1) Click the following link to VIEW the graphical solution; or (2A) highlight and copy the link, then (2B) paste the link into your browser Address bar & press enter:

Use the Backspace key to return to this page:

http://www.solving-math-problems.com/images/rad-eq-2011-08-21-a.png


check the solution by substituting the numerical value of x into the original equation


for x = 4

√(5+x) + √(x) = 5

√(5+4) + √(4) = 5

√(9) + √(4) = 5

3 + 2 = 5, OK → x = 4 is a valid solution



Thanks for writing.

Staff
www.solving-math-problems.com



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