Solve. x^ 4 + 1 = 7x^ 2?

Question

x = (smallest value)

x =

x =

x = (largest value)

Answer 1

x^4 + 1 = 7x^2

Shifting all terms to the LHS......
x^4 - 7x^2 + 1 = 0

Let y = x^2
y^2 - 7y + 1 = 0

Applying quadratic formula......
y = {-(-7) ± sqrt[(-7)^2 - 4(1)(1)]}/[2(1)]
y = [7 ± sqrt(45)]/2

Solving for y......
y = 6.854101966 or y = 0.145898033

Substitute y = x^2......
x^2 = 6.854101966 or x^2 = 0.145898033

Solving for x......
x = 2.618033989 or x = -2.618033989 or
x = 0.38196601 or x = -0.38196601

Therefore, the solutions to the quartic equation are
x = -2.618033989
x = -0.38196601
x = 0.38196601
x = 2.618033989.

Answer 2

Substitue X^2 = Y
Then solve Y^2+1 = 7Y
You will find two values for Y. From there you can find the values of X by taking +-sqrt(Y)