SOLUTION: I'm very confused on the substitution method, so I will write the question as it appears on the assignment. Solve the following equation using the substitution method. {{{x^4-5x^

Algebra ->  Expressions-with-variables -> SOLUTION: I'm very confused on the substitution method, so I will write the question as it appears on the assignment. Solve the following equation using the substitution method. {{{x^4-5x^      Log On


   

Question 737342: I'm very confused on the substitution method, so I will write the question as it appears on the assignment.
Solve the following equation using the substitution method.
x%5E4-5x%5E2-14=0
Complete the following steps to solve the above equation:
a. Let u=x^2, substitute the variables, and write the new equation.
b. Now factor the new equation, apply the zero-product principle, & solve for u.
c. Now substitute u=x^2 to solve for x.
Found 2 solutions by Susan-math, josgarithmetic:
Answer by Susan-math(40) About Me  (Show Source):
You can put this solution on YOUR website!
Sinceu=x%5E2 then u%5E2=%28x%5E2%29%5E2=x%5E4
So x%5E4+-5x%5E2+-14=0 is the same as u%5E2+-+5u+-14+=+0. Which easily factors to be
%28u+-+7%29%28u+%2B+2%29+=+0
Set each factor to zero to solve.
u-7=0++or+u%2B2+=0
This gives us u = 7 or u = -2.
But we don't want u, we want x. Time to substitute usingu=x%5E2
x%5E2=7+or+x%5E2=-2
So we have x=sqrt%287%29x=sqrt%287%29,x=i%2Asqrt%282%29,x=-i%2Asqrt%282%29

Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
Just do exactly as is directed; do what the instructions tell you. Just that simple.

GIVEN: x%5E4-5x%5E2-14=0

a. Let u=x^2, substitute the variables, and write the new equation.
So, do that: u%5E2-5u-14=0.

b. Now factor the new equation, apply the zero-product principle, & solve for u.

You are told to factor the polynomial expression containing u.
%28u+%2B2%29%28u+-7%29=0
One of those binomials or the other must be zero, so u has two solutions.
If u%2B2=0 then u=-2.
If u-7=0 then u=%2B7.


c. Now substitute u=x^2 to solve for x.
Solving for u is part of the solution process but not yet at the finish of the solution. You want solutions for x. You remember, the substitution must now be made for u, in the other direction. x^2=u, because that is how you chose u.

From u=-2, x%5E2=-2. If you do not yet know imaginary numbers, then this means no real solution for x. If you DO know about imaginary numbers, then x=%2B-+sqrt%28-2%29; or x=-i%2Asqrt%282%29 OR x=i%2Asqrt%282%29.

From u=%2B7, x%5E2=7. x=-sqrt%287%29 OR x=sqrt%287%29.

SUMMARY OF THE SOLUTIONS:
highlight%28x=-i%2Asqrt%282%29%29 OR highlight%28x=i%2Asqrt%282%29%29 OR highlight%28x=-sqrt%287%29%29 OR highlight%28x=sqrt%287%29%29.