Solve for x

[Casey Daniels]

Okay trying to figure out if I did this correctly. It wants me to solve for X.

ax⁴ - 4ax² = 0

1. I factored out ax² from the left side.

ax² (x² - 4) = 0

2. Factored the x² - 4 using the difference of perfect square rules

ax²(x - 2) (x + 2) = 0

3. Turned it into 3 different Problems since each section if its equal to 0 would make the whole Left side equal to 0

ax² = 0, (x-2) = 0, (x+2) = 0

4. I quickly see x = 2 or -2. So leaves to solve ax² =0

5. Divide by sides by a

ax² / a = 0 / a

6. Which should leave me with

x² = 0

7. Which leaves me with X=0

8. Final Answer would then be X = 0,2,-2

 

[tkhunny]

Perfect. Do you doubt your excellent work?

 

[JeffM]

Okay trying to figure out if I did this correctly. It wants me to solve for X.

ax⁴ - 4ax² = 0

1. I factored out ax² from the left side.

ax² (x² - 4) = 0

2. Factored the x² - 4 using the difference of perfect square rules

ax²(x - 2) (x + 2) = 0

3. Turned it into 3 different Problems since if any factor is equal to 0 it would make the whole Left side equal to 0

ax² = 0, (x-2) = 0, (x+2) = 0

4. I quickly see x = 2 or -2. So leaves to solve ax² =0

5. Divide by sides by a

ax² / a = 0 / a Only if \(\displaystyle a \ne 0.\)

6. Which should leave me with

x² = 0

7. Which leaves me with X=0

8. Final Answer would then be X = 0,2,-2

All this is very, very good with one exception. Does the problem specify that a is not equal 0? If a = 0, there is no unique solution.

 

[Casey Daniels]

Perfect. Do you doubt your excellent work?

Haven't done math in years, and I usually get a little uncomfortable in the book I'm working through on the challenge problems that have you do stuff with the some unknown letter (like a in this case) instead of dealing with actual numbers.

 

[Casey Daniels]

All this is very, very good with one exception. Does the problem specify that a is not equal 0? If a = 0, there is no unique solution.

That problem they didn't state the a not equal 0 but the previous one they did so I'm assuming they meant it for that one also.

 

[JeffM]

That problem they didn't state the a not equal 0 but the previous one they did so I'm assuming they meant it for that one also.

On the basis of that assumption, your analysis was perfect. Like tkhunny, I am not sure why you doubted yourself. If you had inserted your solutions into the equation, you could have confirmed it yourself.