Algebra problem on final step

  • Thread starter Thread starter Restless
  • Start date Start date
  • Tags Tags
    Algebra Final
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
8 replies · 2K views
Restless
Messages
4
Reaction score
0

Homework Statement


[tex]8x^2-7a^2=19a^2-5ax[/tex]

Homework Equations


None

The Attempt at a Solution


[tex]8x^2-7a^2=19a^2-5ax[/tex]
[tex]13ax^2-7a^2=19a^2[/tex]
[tex]13ax^2=26a^2[/tex]
Now I'm not sure what I should do next...
Maybe divide both sides by 2 to get rid of the squares
[tex]13ax=26a[/tex]
Then say that [itex]ax = 13a[/itex] ?
I am confused.
 
Physics news on Phys.org
Restless said:

Homework Statement


[tex]8x^2-7a^2=19a^2-5ax[/tex]

Homework Equations


None

The Attempt at a Solution


[tex]8x^2-7a^2=19a^2-5ax[/tex]
[tex]13ax^2-7a^2=19a^2[/tex]
[tex]13ax^2=26a^2[/tex]
Now I'm not sure what I should do next...
Maybe divide both sides by 2 to get rid of the squares
[tex]13ax=26a[/tex]
Then say that [itex]ax = 13a[/itex] ?
I am confused.

In your attempt at a solution where did the -5ax term go?

It seems you just dropped it. Oh I see no you can't add the x^2 and x terms together to get 13ax^2.

The strategy should be to group terms so that the right hand side of the equation is a quadratic in x and the left hand side is 0.

then find the roots of the quadratic.
 
Restless said:

Homework Statement


[tex]8x^2-7a^2=19a^2-5ax[/tex]
Is there something you're supposed to do with this, like, say, solve for x?
Restless said:

Homework Equations


None

The Attempt at a Solution


[tex]8x^2-7a^2=19a^2-5ax[/tex]
[tex]13ax^2-7a^2=19a^2[/tex]
You have a mistake above. Apparently you added 5ax to 8x2 and got 13ax2. They are not like terms, so can't be combined.
Restless said:
[tex]13ax^2=26a^2[/tex]
Now I'm not sure what I should do next...
Maybe divide both sides by 2 to get rid of the squares
?
Dividing by 2 doesn't get rid of exponents.
Restless said:
[tex]13ax=26a[/tex]
Then say that [itex]ax = 13a[/itex] ?
There's an error above, as well. You divided the left side by 13, but divided the right side by 2. You can't do that.
Restless said:
I am confused.
 
Ok so I thought about the problem again and now I have

[tex]8x^2-7a^2=19a^2-5ax[/tex]

[tex]8x^2=26a^2-5ax[/tex]

But now I don't see any like terms or anything I can reduce or eliminate
 
Restless said:
Ok so I thought about the problem again and now I have

[tex]8x^2-7a^2=19a^2-5ax[/tex]

[tex]8x^2=26a^2-5ax[/tex]

But now I don't see any like terms or anything I can reduce or eliminate

Bring it to the format
1c110885bd9155bea6b6630e7d24d6c4.png
and
just use the quadratic root equation

3ea647783b5121989cd87ca3bb558916.png


in your case it would be [itex]8 x^2 - 26a^2 + 5ax=0[/itex]
 
Last edited:
I don't understand how to enter my equation into that formula
 
Restless said:
I don't understand how to enter my equation into that formula
in you equation [tex]x=x[/tex]
[tex]a= 8[/tex]
[tex]b= +5a[/tex]
[tex]c= -26a ^ 2[/tex]
 
Hysteria X said:
Bring it to the format
1c110885bd9155bea6b6630e7d24d6c4.png
and
just use the quadratic root equation

3ea647783b5121989cd87ca3bb558916.png


in your case it would be [itex]8 x^2 - 26a^2 + 5ax=0[/itex]

It would be better as 8x2 + 5ax - 26a2 = 0
 
Does it tell you to solve for X, it's a bit confusing, since you have left it unmentioned or I am missing something. Anywho, what you cannot do for certain is divide the sides of the equal sign with x or a. If it were x you were solving for, if you divide it like that, then one of your solutions will disappear, the number of the dimension of the variable tells you how many solutions there will have to be.

You can treat both of the numbers a and x as variables here. 1 time, solve for x and then solve again for a, since you haven't said what needs to be done. I have a hunch one needs to solve for x here.