Solving Quadratic Equations by the Quadratic Formula

Hey Guys,

I am a bit confused on what to do for this problem. If I can get some help I'd appreciate it.

Suppose that f(x) = x^{4 }+ 7x^{2}+ 12. Find the values of x such that

a. f(x) = 12

b. f(x) = 6

I figured part b already but I am having trouble on part (a) which is

12 = x^{4 }+ 7x^{2}+ 12

-12 -12

Now I'm left with x^{4 + }7x^{2 }

I use "Let u = x^{2} , Let u^{2 = }x^{4}^{"}

u^{2 + }u

Do I get a third term by completing the square method? I am lost with this step.

Re: Solving Quadratic Equations by the Quadratic Formula

It is not must that quadratic polynomial should always have three terms. It may be a monomial / binomial or a trinomial.

Now you have u(u+ 7u ) = 0

In second case we will have

u^2 + 7u + 6 = 0

Re: Solving Quadratic Equations by the Quadratic Formula

What is the quadratic formula? What are the values of a, b, and c?

Re: Solving Quadratic Equations by the Quadratic Formula

Quote:

Originally Posted by

**ibdutt** It is not must that quadratic polynomial should always have three terms. It may be a monomial / binomial or a trinomial.

Now you have u(u+ 7u ) = 0

In second case we will have

u^2 + 7u + 6 = 0

Hey ibdutt,

You are correct :D

Honestly, I did factor the problem and ended up with u(u + 7u) = 0 but I thought it looked wrong and I wasn't used to seeing a quadratic polynomial with two terms. Thanks for clearing things up for me!

Cheers!

Re: Solving Quadratic Equations by the Quadratic Formula

Quote:

Originally Posted by

**mathguy25** What is the quadratic formula? What are the values of a, b, and c?

As ibdutt mentioned "It is not must that quadratic polynomial should always have three terms. It may be a monomial / binomial or a trinomial." :)