Intermediate Algebra: solve (2m - 1 ) ^(2) = -25

[Helen]

Can I have some help with this?
Solve the equation. Express radicals in simplest form.
(2m - 1 ) ^(2) = -25
I have worked on this in so many ways.
2m - 1 = radical 25 or 2m - 1 = - radical 25
2m - 1 = 5 or 2m - 1 = - radical 5
2m = 5 or 2m = 6
m = 5/2 or m = - 3
solution set: {-3, 5/2}
I rarely get things right on the first couple of tries, so I don't think that this is an exception.

 

[o_O]

Re: Intermediate Algebra

First off, does your original question have the -25 or did you simply make a typo? If it does have a -25, then m has no real solutions since you can't take the square roots of negative numbers.

If the question looked like (2m - 1)[sup:31yyyplt]2[/sup:31yyyplt] = +25, then you made a few calculation errors:

2m - 1 = radical 25 or 2m - 1 = - radical 25
2m - 1 = 5 or 2m - 1 = - radical 5 I think you accidentally typed in the 'radical'
2m = 5 or 2m = 6 What happened to the -1 for (2m - 1 = 5)? And for (2m - 1 = -5), what happens when you add +1 to both sides?
m = 5/2 or m = - 3 Fix your mistakes on the previous line should fix your problem
solution set: {-3, 5/2}

 

[jwpaine]

Re: Intermediate Algebra

Can I have some help with this?
Solve the equation. Express radicals in simplest form.
(2m - 1 ) ^(2) = -25
I have worked on this in so many ways.
2m - 1 = radical 25 or 2m - 1 = - radical 25
2m - 1 = 5 or 2m - 1 = - radical 5
2m = 5 or 2m = 6
m = 5/2 or m = - 3
solution set: {-3, 5/2}
I rarely get things right on the first couple of tries, so I don't think that this is an exception.

\(\displaystyle (2m - 1 ) ^2 = -25\)

square root both sides
|2m - 1| = 5i

.....etc

Your solutions will be imaginary because you can't take the square root of a negative number.
John

 

[Loren]

Re: Intermediate Algebra

Something's wrong. I think you have copied the problem incorrectly. The solution set you give does not satisfy the given equation.

(2m - 1 ) ^(2) = -25
I would expand the left side, then add 25 to both sides. That should yield something like 4m^2 - 4m + 26 = 0. I would try to factor the left side. If that doesn't work I guess I would use the quadratic formula.

If you start out by taking the square root of both sides you will get \(\displaystyle 2m-1=\pm \sqrt{-25}\). This leads to the solution set \(\displaystyle m=\frac{1\pm 5i}{2}\).

 

[Helen]

Re: Intermediate Algebra

o_0,
Could the answer be {1 + 5i /2, 1 - 5i /2}
Thank you for your help. Helen

 

[o_O]

Re: Intermediate Algebra

Well, that looks like Loren's answer so ... I'll assume it's right :wink:

And sorry, I didn't realize you dealt with imaginary numbers.

 

[jwpaine]

Re: Intermediate Algebra

I would expand the left side, then add 25 to both sides. That should yield something like 4m^2 - 4m + 26 = 0. I would try to factor the left side. If that doesn't work I guess I would use the quadratic formula.

4m^2 - 4m + 26 = 0

Will be solved to give you the same imaginary roots as

|2m - 1| = 5i

As the above can be squared back into it's original form. (|2m - 1|)^2 = (2m - 1)^2 and (5i)^2 = -25.
What you do to one side, you must to do the other, thus they are both the same. Trying to factor it more when one side was already factored, is redundant.

John

 

[Loren]

Re: Intermediate Algebra

Helen. While you are at it, you might consider learning to use grouping symbols and also learn the order of operations convention.

Your response of "Could the answer be {1 + 5i /2, 1 - 5i /2}" means you think the roots are \(\displaystyle 1 \pm \frac{5i}{2}\) which is not correct. You should have typed in "{(1 + 5i)/2, (1 - 5i)/2}". See the difference?

 

[Mrspi]

Can I have some help with this?
Solve the equation. Express radicals in simplest form.
(2m - 1 ) ^(2) = -25
I have worked on this in so many ways.
2m - 1 = radical 25 or 2m - 1 = - radical 25
2m - 1 = 5 or 2m - 1 = - radical 5
2m = 5 or 2m = 6
m = 5/2 or m = - 3
solution set: {-3, 5/2}
I rarely get things right on the first couple of tries, so I don't think that this is an exception.

You have

(2m - 1)[sup:1uxuoqob]2[/sup:1uxuoqob] = -25

Take the square root of both sides of the equation. Remember that on the right side, you'll have two square roots, + and -:

sqrt[ (2m - 1)[sup:1uxuoqob]2[/sup:1uxuoqob]] = +/- sqrt(-25)

2m - 1 = +/- 5i

Add 1 to both sides:

2m - 1 + 1 = 1 +/- 5i
2m = 1 +/- 5i

Divide both sides by 2:

m = (1 +/- 5i) / 2

These are not REAL solutions, but they are solutions in the complex numbers.

 

[Helen]

, Loren, jwpaine and mrspi,
I want to thank all of you for the help that you have given.
, I am just getting into the imaginary numbers. I always appreciate your time.
Loren, I do see the difference and will keep trying.
Thank you all again. Helen