SOLUTION: Can you please help me solve this problem?!!? {{{9x^4=25x^2-16}}} The directions say: solve the equation by making the appropriate substitution. Here is what I have so far....b

Algebra ->  Expressions-with-variables -> SOLUTION: Can you please help me solve this problem?!!? {{{9x^4=25x^2-16}}} The directions say: solve the equation by making the appropriate substitution. Here is what I have so far....b      Log On


   

Question 180326: Can you please help me solve this problem?!!? 9x%5E4=25x%5E2-16
The directions say: solve the equation by making the appropriate substitution.
Here is what I have so far....but I am stuck.
9x%5E4-25x%5E2%2B16=0
%283x%5E2%29%5E2-25x%5E2%2B16=0
3u%5E2-25u%2B16=0
At this point I wasn't sure. Do I use the quadratic formula??? I tried that but just got confused. Please help. Thanks.
Found 3 solutions by Alan3354, jim_thompson5910, stanbon:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
9x%5E4-25x%5E2%2B16=0
%283x%5E2%29%5E2-25x%5E2%2B16=0
3u%5E2-25u%2B16=0
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Sub u for x^2
9u%5E2+-+25u+%2B+16+=+0 You used 3u for one term, and u for the other.
You can factor"
(9u-16)*(u-1) = 0
u = 1, u = 16/9
Then
x = +1, -1
x = +4/3, -4/3
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Another approach is to "eyeball" the 9, -25, 16 and spot that +1 and -1 are solutions, factor those out, then you have a quadratic.

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
9x%5E4-25x%5E2%2B16=0 Start with the given equation.

Let u=x%5E2. This means that u%5E2=x%5E4


9u%5E2-25u%2B16=0 Replace x%5E4 with u%5E2 and x%5E2 with u


Notice we have a quadratic equation in the form of au%5E2%2Bbu%2Bc where a=9, b=-25, and c=16


Let's use the quadratic formula to solve for u


u+=+%28-b+%2B-+sqrt%28+b%5E2-4ac+%29%29%2F%282a%29 Start with the quadratic formula


u+=+%28-%28-25%29+%2B-+sqrt%28+%28-25%29%5E2-4%289%29%2816%29+%29%29%2F%282%289%29%29 Plug in a=9, b=-25, and c=16


u+=+%2825+%2B-+sqrt%28+%28-25%29%5E2-4%289%29%2816%29+%29%29%2F%282%289%29%29 Negate -25 to get 25.


u+=+%2825+%2B-+sqrt%28+625-4%289%29%2816%29+%29%29%2F%282%289%29%29 Square -25 to get 625.


u+=+%2825+%2B-+sqrt%28+625-576+%29%29%2F%282%289%29%29 Multiply 4%289%29%2816%29 to get 576


u+=+%2825+%2B-+sqrt%28+49+%29%29%2F%282%289%29%29 Subtract 576 from 625 to get 49


u+=+%2825+%2B-+sqrt%28+49+%29%29%2F%2818%29 Multiply 2 and 9 to get 18.


u+=+%2825+%2B-+7%29%2F%2818%29 Take the square root of 49 to get 7.


u+=+%2825+%2B+7%29%2F%2818%29 or u+=+%2825+-+7%29%2F%2818%29 Break up the expression.


u+=+%2832%29%2F%2818%29 or u+=++%2818%29%2F%2818%29 Combine like terms.


u+=+16%2F9 or u+=+1 Simplify.


---------------------------------------

Now remember, we let u=x%5E2, so this means that


x%5E2+=+16%2F9 or x%5E2+=+1



Let's solve the first equation x%5E2+=+16%2F9:


x+=+0%2B-sqrt%2816%2F9%29 Take the square root of both sides


x+=+sqrt%2816%2F9%29 or x+=+-sqrt%2816%2F9%29 Break up the "plus/minus"


x+=+4%2F3 or x+=+-4%2F3 Take the square root of 16%2F9 to get 4%2F3



So the first two solutions are x+=+4%2F3 or x+=+-4%2F3


------------------------------




Now let's solve the second equation x%5E2+=+1:


x+=+0%2B-sqrt%281%29 Take the square root of both sides


x+=+sqrt%281%29 or x+=+-sqrt%281%29 Break up the "plus/minus"


x+=+1 or x+=+-1 Take the square root of 1 to get 1



So the next two solutions are x+=+1 or x+=+-1


======================================================

Answer:


So the 4 solutions are:


x+=+4%2F3, x+=+-4%2F3, x+=+1 or x+=+-1

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
9x^4=25x^2-16
9x^4 - 25x^2 + 16 = 0
Quadratic Formula:
x^2 = [25 +- sqrt(625 - 4*9*+16)]/18
x^2 = [25 +- sqrt(49)]/18
x^2 = [25 + 7]/18 or x^2 = [25 - 7]/18
x^2 = 16/9 or x^2 = 1
x = (4/3) or x = (-4/3) or x = 1 or x= -1
===============================================
Cheers
Stan H.