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 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.