Principle of Square Roots: Solve 5x^2=125, 7x^2-9=0

juliemariefort

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Apr 3, 2009
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Hello. Are these right?

#1) Solve 5x^2=125.

x^2=25
x= squareroot 25 or x= negative squareroot 25
so, x=5 or x= -5

#2) Solve 7x^2-9=0

x^2= 9/7
x= squareroot 9/7 or x= negative squareroot 9/7

Thanking you in advance.
 


Here is how we type square roots.

x = sqrt(5)

x = -sqrt(5)

In other words, the number under the square root sign goes inside the parentheses.

Do you know? You can check your own answers.

If you want to know whether or not the expression 5x^2 equals 125 when x is sqrt(5), then substitute x = sqrt(5) into the expression and evaluate.

5 * [sqrt(5)]^2

If that equals 125, then x = sqrt(5) is a solution.

Do the same to check x = -sqrt(5).

Likewise, with your solutions for 7x^2 - 9 = 0.

Substitute sqrt(9/7) for x, do the arithmetic, and see if you end up with zero.

Check x = -sqrt(9/7), too.

By the way, if you've studied how to simplify radicals, then your instructor probably does not want to see sqrt(9/7) because this expression can be simplified.

x = 3/sqrt(7)

Your instructor might also expect to see you rationalize the denominator, if you've studied that, too.

x = 3 sqrt(7)/7

Please show your work if you need more help, and try to say something about why you're stuck, so that I might determine where to continue helping you.

If I typed anything that you do not understand, then please ask specific questions.

 
juliemariefort said:
Hello. Are these right?

#1) Solve 5x^2=125.

x^2=25
x= squareroot 25 or x= negative squareroot 25
so, x=5 or x= -5

#2) Solve 7x^2-9=0

x^2= 9/7
x= squareroot 9/7 or x= negative squareroot 9/7

Thanking you in advance.

Yes, perfect.

Note that \(\displaystyle \sqrt{\frac{9}{7}}=\frac{3}{\sqrt{7}}\)
 
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