SOLUTION: Solve. 6x^4-19x^2+10=0 I tried using the solver and the answer didn't look right to me. Can you show me how to solve so I can understand the steps.

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Solve. 6x^4-19x^2+10=0 I tried using the solver and the answer didn't look right to me. Can you show me how to solve so I can understand the steps.      Log On


   

Question 132426: Solve.
6x^4-19x^2+10=0
I tried using the solver and the answer didn't look right to me. Can you show me how to solve so I can understand the steps.
Found 2 solutions by checkley71, jim_thompson5910:
Answer by checkley71(8403) About Me  (Show Source):
You can put this solution on YOUR website!
6x^4-19x^2+10=0
(3x^2-2)(2x^2-5)=0 answer.

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Which solver did you use?

6x%5E4-19x%5E2%2B10=0+ Start with the given equation


Let u=x%5E2

6u%5E2-19u%2B10=0+ Plug in u=x%5E2


Solved by pluggable solver: Quadratic Formula
Let's use the quadratic formula to solve for u:


Starting with the general quadratic


au%5E2%2Bbu%2Bc=0


the general solution using the quadratic equation is:


u+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29




So lets solve 6%2Au%5E2-19%2Au%2B10=0 ( notice a=6, b=-19, and c=10)





u+=+%28--19+%2B-+sqrt%28+%28-19%29%5E2-4%2A6%2A10+%29%29%2F%282%2A6%29 Plug in a=6, b=-19, and c=10




u+=+%2819+%2B-+sqrt%28+%28-19%29%5E2-4%2A6%2A10+%29%29%2F%282%2A6%29 Negate -19 to get 19




u+=+%2819+%2B-+sqrt%28+361-4%2A6%2A10+%29%29%2F%282%2A6%29 Square -19 to get 361 (note: remember when you square -19, you must square the negative as well. This is because %28-19%29%5E2=-19%2A-19=361.)




u+=+%2819+%2B-+sqrt%28+361%2B-240+%29%29%2F%282%2A6%29 Multiply -4%2A10%2A6 to get -240




u+=+%2819+%2B-+sqrt%28+121+%29%29%2F%282%2A6%29 Combine like terms in the radicand (everything under the square root)




u+=+%2819+%2B-+11%29%2F%282%2A6%29 Simplify the square root (note: If you need help with simplifying the square root, check out this solver)




u+=+%2819+%2B-+11%29%2F12 Multiply 2 and 6 to get 12


So now the expression breaks down into two parts


u+=+%2819+%2B+11%29%2F12 or u+=+%2819+-+11%29%2F12


Lets look at the first part:


x=%2819+%2B+11%29%2F12


u=30%2F12 Add the terms in the numerator

u=5%2F2 Divide


So one answer is

u=5%2F2




Now lets look at the second part:


x=%2819+-+11%29%2F12


u=8%2F12 Subtract the terms in the numerator

u=2%2F3 Divide


So another answer is

u=2%2F3


So our solutions are:

u=5%2F2 or u=2%2F3





Remember, we let u=x%5E2. So this means that


x%5E2=5%2F2 or x%5E2=2%2F3


x=0%2B-sqrt%285%2F2%29 or x=0%2B-sqrt%282%2F3%29 Take the square root of both sides for each case


Break up the two expressions

x=sqrt%285%2F2%29, x=-sqrt%285%2F2%29, x=sqrt%282%2F3%29 or x=-sqrt%282%2F3%29



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

Answer:


So our solutions are

x=sqrt%285%2F2%29, x=-sqrt%285%2F2%29, x=sqrt%282%2F3%29 or x=-sqrt%282%2F3%29