Algebra problem on final step

In summary, the student is trying to solve an equation 8x^2-7a^2=19a^2-5ax. They have a mistake where they added 5ax to 8x2, and then found that 13ax^2=26a^2. They are now not sure what to do next and may divide by 2 to get rid of exponents. However, this does not work, because 13ax=26a. Then they say that ax=13a. They are now confused.
  • #1
Restless
4
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.
 
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  • #2
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.
 
  • #3
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.
 
  • #4
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
 
  • #5
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:
  • #6
I don't understand how to enter my equation into that formula
 
  • #7
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]
 
  • #8
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
 
  • #9
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.
 

FAQ: Algebra problem on final step

1. What is the final step in solving an algebra problem?

The final step in solving an algebra problem is to check your answer for accuracy and completeness. This involves plugging your solution back into the original equation and making sure it satisfies all the given conditions.

2. How do I know if I have solved an algebra problem correctly?

You can check your solution by substituting it back into the original equation and simplifying both sides. If both sides of the equation are equal, then you have solved the problem correctly.

3. What should I do if I get a different answer than the one provided in the answer key?

If your answer differs from the one provided in the answer key, double check your work and make sure you followed all the steps correctly. If you are still unsure, consult with your teacher or a classmate for clarification.

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Yes, you can use a calculator to help you solve algebra problems. However, it is important to understand the concepts and steps involved in solving the problem by hand before relying on a calculator.

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There are some common patterns and tricks that can be used to solve certain types of algebra problems. However, it is important to understand the underlying concepts and steps involved in solving the problem in order to use these shortcuts effectively.

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