Finding Solutions

[adr8]

We are asked to find all possible solutions, by using any method.

2x^2-5=0

I tried factoring this one out but I noticed that I couldnt factor it out. Anyways, this is what I did though I moved the five to the other side so I have:

2x^2=5

Then I moved the 2 to the other side by cancelling it out:

x^2=5/2


Then to cancel the exponent I used the square root so my answer is

x=Square root (5/2)


My question is did I do this right? In other words, is this my only solution for this problem?


I feel confident that I have it right, but I just want to make sure that I didnt skip anything.

 

[mmm4444bot]

find all possible solutions, by using any method.

2x^2 - 5 = 0

I tried factoring this one out but I noticed that I couldnt factor it out.

Actually, we can factor it, if we think "difference of squares".

This requires the realization that the constant 2 is:

\(\displaystyle \big ( \sqrt{2} \big )^{2}\)

and the constant 5 is:

\(\displaystyle \big ( \sqrt{5} \big )^{2}\)

Therefore, we can write:

\(\displaystyle 2x^2 - 5 = (\sqrt{2}x)^2 - (\sqrt{5})^2\)

This is a difference of squares; it factors as such:

\(\displaystyle \left ( \sqrt{2}x + \sqrt{5} \right ) \left ( \sqrt{2}x - \sqrt{5} \right )\)

Use the Zero-Product Property to solve for x. That is, set each factor equal to zero, and solve for x.

I'll do the first one:

\(\displaystyle \sqrt{2}x + \sqrt{5} = 0\)

\(\displaystyle \sqrt{2}x = - \sqrt{5}\)

\(\displaystyle x = - \frac{\sqrt{5}}{\sqrt{2}}\)

\(\displaystyle x = -\sqrt{\frac{5}{2}}\)

You did it correctly using a different method, but as Jeff said:

Solving an equation of the form \(\displaystyle x^2 = C\) means \(\displaystyle x = \pm \sqrt{C}\)

Also, if you have not yet learned about fractional exponents (eg: x^1/2), don't be concerned. You will learn later that x^1/2 is just another way (the exponential way) of writing sqrt(x).

 

[adr8]

Good to hear I have it right . Thanks....

 

[mmm4444bot]

Oh, I meant that your reasoning was correct, but I should not have implied that your answer is correct because you missed the negative radical. But, you understand now, yes?