Solve 15x^4 - 13x^2 + 2 = 0

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Gr8fu13

Junior Member
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Feb 13, 2011
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The equation to solve is:
15x^4 - 13x^2 + 2 = 0

I let u=x^2

This gave me 15u^2 - 13x + 2 = 0
I tried to factor but couldn't get past this:
(3u )(3u )

But I could not come up with answers that have a sum of -19 and a product of 2.
Can someone please help me with this.
I also tried the quadratic equation but it did not seem to help either.
 
Keep trying, Let's see your Quadratic Equation efforts.

13^2 - 4*15*2 = 169 - 120 = 49 = 7^2 -- You should be able to factor it,

Give it another go.
 
I think I got it. Can someone please confirm?
15x^4 - 13x^2 + 2 = 0
Let u=x^2
15x^2 - 13x + 2 = 0

Factoring:
(3u-1)(5u-2)
x=1/3 and x=2/5

So the answer would be:
-?3/3,?3/3,-?10/5,?10/5

Would that be correct?
 
Gr8fu13 said:
I think I got it. Can someone please confirm?
15x^4 - 13x^2 + 2 = 0
Let u=x^2
15x^2 - 13x + 2 = 0

Factoring:
(3u-1)(5u-2) = 15u[sup:43acomi9]2[/sup:43acomi9] - 11u + 2 .... not the equation you tried to factorize!!

x=1/3 and x=2/5

So the answer would be:
-?3/3,?3/3,-?10/5,?10/5

Would that be correct?
Click to expand...
 
Is this equation unsolvable? I have racked my brain for other numbers and I am not getting it. Maybe I have been looking at the problem too long. It's probably real simple and I'm just not seeing it.
 
Gr8fu13 said:
I think I got it. Can someone please confirm? Would that be correct?
Click to expand...

Hmmm... This is not encouraging. You MUST gain a little confidence and some ability to check your own work. Really, would it have been so hard to multiply the binomials and see that you managed 11 and not 13? You should be able to do that.

15 = 1*15
15 = 3*5

2 = 1*2

There aren't very many choices. Try them ALL until you find it. We already proved it could be factored.
 
15x^4 - 13x^2 + 2 = 0
Let u=x^2
15x^2 - 13x + 2 = 0

Factoring:
(1u-2)(15u-1)

x=2/1 and x=1/15
So the answer would be:
-?2/1,?2/1,-?15/15,?15/15
Or would I simplify it to:
-?2,?2,-?15/15,?15/15
I'm guessing the first of the two would be correct.
I knew it was easy and in front of my face. I just kept thinking that 3 * 5 is the only solution to make 15.
 
Gr8fu13 said:
15x^4 - 13x^2 + 2 = 0
Let u=x^2
15x^2 - 13x + 2 = 0
Click to expand...

Factoring:
(1u-2)(15u-1)
Click to expand...

If you FOIL this you get \(\displaystyle 15u^{2}-u-30u+2=15u^{2}-31u+2=0\)

This is not the problem you started with. So, the factoring must be incorrect.

x=2/1 and x=1/15
So the answer would be:
-?2/1,?2/1,-?15/15,?15/15
Or would I simplify it to:
-?2,?2,-?15/15,?15/15
I'm guessing the first of the two would be correct.
I knew it was easy and in front of my face. I just kept thinking that 3 * 5 is the only solution to make 15.
Click to expand...

Sorry, not the correct solutions.

What two numbers multiplied equal 30 and when added equal -13?.

You seem to be hung up on 15 and 2, or some variation thereof.

How about -10 and -3?. Now proceed.

If you make a u substitution, don't forget to resub.
 
I appreciate all your patience through this. I just really need to understand this for my final exam.
I don't understand how we looking for 30? I thought the numbers were suppose to equal 2 when multiplied and -13 when added.
So would it be something like
(1u-3)(15u-10) :?:
 
Look buddy, yer running around in a dark tunnel!

15u^2 - 13u + 2 = 0
Start by solving that using the QUADRATIC formula....what do you get?
 
For reference,

if Ax[sup:2q5q8z8v]2[/sup:2q5q8z8v] + Bx + C = 0

then

(x - x[sub:2q5q8z8v]1[/sub:2q5q8z8v])(x - x[sub:2q5q8z8v]2[/sub:2q5q8z8v]) = 0

where:

x[sub:2q5q8z8v]1,2[/sub:2q5q8z8v] = [-B ± ?(B[sup:2q5q8z8v]2[/sup:2q5q8z8v] - 4*A*C)]/(2*A)

In your case

A = 15, B = -13 and C = 2

Now go for it......

.
 
Gr8fu13 said:
I appreciate all your patience through this. I just really need to understand this for my final exam.
I don't understand how we looking for 30? I thought the numbers were suppose to equal 2 when multiplied and -13 when added.
So would it be something like
(1u-3)(15u-10) :?:
Click to expand...

You multiply the constant by the coefficient of the x^2 term.

2 would be OK if it were just x^2 and not 15x^2.

What two numbers when multiplied equal "the constant multiplied by the x^2 coefficient", and when added equal the coefficient of the linear term.

Say you have a quadratic like \(\displaystyle x^{2}+9x+18\), and you want to factor.

What two numbers when multiplied equal 18*1=18. The coefficient of the x^2 term is 1. This is why you use two numbers multiplied equal 18.

See now?. If there is a coefficient other than 1, you will have another number besides the constant. In this case 15*2=30.

\(\displaystyle 15u^{2}-10u-3u+2\)

Now, group and factor. Remember to resub when you're finished.

But, as Denis and SK suggested, you can always use the quad formula if you are not required to factor.
 
Would this work?
5x^2(3x^2 - 2) -1(3x^2 - 2) = 0

(3x^2 - 2) (5x^2 - 1) = 0

3x^2 - 2 = 0 or 5x^2 - 1 = 0

x^2 = 2/3 or x^2 = 1/5

x = +/- sqrt(2/3) or x = +/- sqrt(1/5)

x^2 = 2/3 or x^2=1/5
x=-(sqrt6)/3, sqrt6)/3,-(sqrt5)/5,(sqrt5)/5
 
Gr8fu13 said:
Would this work? 5x^2(3x^2 - 2) -1(3x^2 - 2) = 0

(3x^2 - 2) (5x^2 - 1) = 0

3x^2 - 2 = 0 or 5x^2 - 1 = 0

x^2 = 2/3 or x^2 = 1/5

x = +/- sqrt(2/3) or x = +/- sqrt(1/5)

x^2 = 2/3 or x^2=1/5
x=-(sqrt6)/3, sqrt6)/3,-(sqrt5)/5,(sqrt5)/5
Click to expand...

There is an easy check!!!

Put those values in your equation and find out whether those satisfy the given equation.

Try it.....
 
hmmm, it doesn't seem to check unless I am doing it wrong. I came up with 216 for 6/3 and 4 for 5/5. I don't get it!! I am about to just get this one wrong on purpose:(
 
Gr8fu13 said:
5x^2(3x^2 - 2) -1(3x^2 - 2) = 0
(3x^2 - 2) (5x^2 - 1) = 0
3x^2 - 2 = 0 or 5x^2 - 1 = 0
x^2 = 2/3 or x^2 = 1/5
x = +/- sqrt(2/3) or x = +/- sqrt(1/5)
Click to expand...
STOP there!!
(what you did after makes no sense...)

Now check by substituting back in original equation; start with x = sqrt(2/3):
15*[sqrt(2/3)]^4 - 13*[sqrt(2/3)]^2 + 2 : does that equal 0? YES it does, so CORRECT!

Do same with other solutions.
 
Thanks Denis! I think I was not resubing the x^2 that I used for u. I hate that, one little mishap and all your work is in vein.
 
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