Algebra problem

Valmont · 27-08-2014 06:29AM #1

I had to look at the solution for a geometry problem and one of the steps is beyond me at the moment, I hope someone can help:

1. SQRT((X/2)^2)+(X/2)^2))

becomes

2. SQRT(2*(X^2/2^2) I understand this step but...

becomes

3. x/SQRT(2)

When I try to simplify step 2 I always get SQRT(2)*X/2 - I don't see how to get to step 3? Thanks in advance and I hope my equations make sense?

Ciaran · 27-08-2014 11:17AM

Valmont wrote: »

I had to look at the solution for a geometry problem and one of the steps is beyond me at the moment, I hope someone can help:

1. SQRT((X/2)^2)+(X/2)^2))

becomes

2. SQRT(2*(X^2/2^2) I understand this step but...

becomes

3. x/SQRT(2)

When I try to simplify step 2 I always get SQRT(2)*X/2 - I don't see how to get to step 3? Thanks in advance and I hope my equations make sense?

sqrt(2)/2 = 1/sqrt(2) because 2 is sqrt(2) squared.

TheBody · 27-08-2014 09:21PM

You have:

[latex]\sqrt{\frac{2x^2}{2^2}}=\frac{\sqrt{2}x}{2} [/latex].

If we multiply top and bottom by [latex]\sqrt{2}[/latex] we get:

[latex]\frac{\sqrt{2}x}{2}\frac{\sqrt{2}}{\sqrt{2}}=\frac{2x}{2\sqrt{2}}=\frac{x}{\sqrt{2}} [/latex].

Simples!

Valmont · 28-08-2014 06:19AM

Thank you both. TheBody, how did you get the graphics in your post?

Kavrocks · 28-08-2014 07:43AM

Valmont wrote: »

Thank you both. TheBody, how did you get the graphics in your post?

He used LaTeX inside [LaTeX] tags. If you try and quote his response you will see what he used.

bnt · 28-08-2014 12:14PM

I tried doing it from the inside out:
SQRT [ (x/2)² +(x/2)² ]
= SQRT [ x²/4 + x²/4 ]
= SQRT [ x²/2 ]
= x / √2

Valmont · 16-09-2014 06:21PM

I have another seemingly insoluble algebra problem. I have to solve the following equation for y in order to graph the line:
[latex]12+5\frac{-y+7}{2}=3y[/latex]
My answer was repeatedly, after many checks and rechecks, [latex]y=\frac{59}{11}[/latex]. After being told it was incorrect, I entered the equation into this online calculator which again gives the answer as [latex]y=\frac{59}{11}[/latex].

Naturally I marked the question as wrong I looked up the solution which has an equally logical but completely different solution. They take [latex]12+5\frac{-y+7}{2}=3y[/latex] and turn it into [latex]24-5y-35=6y[/latex] and I understand how this is done. The only problem is that if we now solve for y we get [latex]y=-1[/latex]! One has to be wrong which means either Khanacademy are wrong or the mathpapa.com online calculator is! Please help!

Ciaran · 16-09-2014 09:54PM

Valmont wrote: »

I have another seemingly insoluble algebra problem. I have to solve the following equation for y in order to graph the line:
[latex]12+5\frac{-y+7}{2}=3y[/latex]
My answer was repeatedly, after many checks and rechecks, [latex]y=\frac{59}{11}[/latex]. After being told it was incorrect, I entered the equation into this online calculator which again gives the answer as [latex]y=\frac{59}{11}[/latex].

Naturally I marked the question as wrong I looked up the solution which has an equally logical but completely different solution. They take [latex]12+5\frac{-y+7}{2}=3y[/latex] and turn it into [latex]24-5y-35=6y[/latex] and I understand how this is done. The only problem is that if we now solve for y we get [latex]y=-1[/latex]! One has to be wrong which means either Khanacademy are wrong or the mathpapa.com online calculator is! Please help!

24-5y-35=6y should be 24-5y+35=6y which gives your original answer.

Algebra problem

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