I don't either!
You have probably looked up the answer for the wrong question.
Also, shouldn't the original question be:
6x^4-7x^2+2=0
instead of
6^4-7x^2+2=0 ?
If that's the case, think of
y=x² and rewrite the question in terms of y.
Question solve
6^4-7x^2+2=0
MY ANSWER
I get 28b^13+b^2+22b+26 over 4b+3
but I do not understand how this can be the solution for x
4 answers
6x^4-7x^2+2=0
is it 2^1/2 over 3^1/2
is it 2^1/2 over 3^1/2
Solve means to find the value of the unknown variable, namely x.
6x^4-7x^2+2=0 can be factorized as
(2x²-1)(3²-2)=0
So you would equate each factor to zero to find the roots (4 roots in all).
For example,
(2x²-1)=0
2x²=1
x²=(1/2)
x=±(&radic(2)/2)
I'll leave it to you to find the other two roots.
6x^4-7x^2+2=0 can be factorized as
(2x²-1)(3²-2)=0
So you would equate each factor to zero to find the roots (4 roots in all).
For example,
(2x²-1)=0
2x²=1
x²=(1/2)
x=±(&radic(2)/2)
I'll leave it to you to find the other two roots.
(2x²-1)(3²-2)=0 should read
(2x²-1)(3x²-2)=0,
and √(2/3) is one of the remaining roots.
(2x²-1)(3x²-2)=0,
and √(2/3) is one of the remaining roots.
0 answers