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
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.
6x^4-7x^2+2=0
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.
(2x²-1)(3²-2)=0 should read
(2x²-1)(3x²-2)=0,
and √(2/3) is one of the remaining roots.
To solve the equation 6^4 - 7x^2 + 2 = 0, we need to follow a series of steps to isolate the variable x and find its values.
Step 1: Start with the given equation: 6^4 - 7x^2 + 2 = 0.
Step 2: Simplify the equation. Calculate 6^4 = 1296 and rewrite the equation as 1296 - 7x^2 + 2 = 0.
Step 3: Combine like terms. The equation now becomes 1298 - 7x^2 = 0.
Step 4: Move the constant term to the other side of the equation. Subtract 1298 from both sides: -7x^2 = -1298.
Step 5: Divide both sides of the equation by -7 to isolate x^2. The equation becomes x^2 = 1298/7.
Step 6: Take the square root of both sides to solve for x. The positive and negative square roots must be considered, so we have: x = ±√(1298/7).
This is the general solution for x, represented by the square root of 1298/7. However, the expression you provided as your answer, 28b^13+b^2+22b+26 over 4b+3, seems unrelated to the given equation and its solution. It appears to be a different expression involving the variable b, not x.