Finding the real-number solutions of an equation

[Kiyabear]

I don't really understand how to find the real-number solutions of the problem: 7x^4 = 252x^2
Please help

 

[Dr.Peterson]

Rewrite it with 0 on one side, then factor out the GCF. Show us what you get, and we can help you with the next step if you don't already see it.

 

[Steven G]

You should at least see that x=0 is a solution.

 

[pka]

I don't really understand how to find the real-number solutions of the problem: 7x^4 = 252x^2

See the solution HERE.

 

[Kiyabear]

Rewrite it with 0 on one side, then factor out the GCF. Show us what you get, and we can help you with the next step if you don't already see it.

Thank you so much! That helped a lot.

 

[HallsofIvy]

Yes, that works but personally, I would not write the equation as \(\displaystyle 7x^4- 252x^2= 0\) because I believe that \(\displaystyle 7x^4= 252x^2\) is already simpler than that. I would observe, as Jomo says, that 0 is a solution. If x is NOT 0, then we can divide both sides by \(\displaystyle x^2\) to get \(\displaystyle 7x^2= 252\) and then divide both sides by 7: \(\displaystyle x^2= 252/7= 36= 6^2\).

 

[Steven G]

Yes, that works but personally, I would not write the equation as \(\displaystyle 7x^4- 252x^2= 0\) because I believe that \(\displaystyle 7x^4= 252x^2\) is already simpler than that. I would observe, as Jomo says, that 0 is a solution. If x is NOT 0, then we can divide both sides by \(\displaystyle x^2\) to get \(\displaystyle 7x^2= 252\) and then divide both sides by 7: \(\displaystyle x^2= 252/7= 36= 6^2\).

To OP. I too agree with HallsofIvy that it would be best not to set this equation equal to 0 as long as you note that x=0 is a solution. Dr Peterson told you to set it equal to 0 as this is the standard way of solving problems of this type. However this particular problems just begs you not to do that.