How do you find the roots for $-x^2+4x-8$?

Question

I am having trouble finding the roots for $-x^2+4x-8$. I tried factoring it regularly and tried the quadratic formula and neither of them worked. Could someone please help me out here? Thanks in advance.

Edit: Apparently the quadratic formula does in fact work. When I tried it, the square root number was negative and as a result it gave me an error so I don't know what to do from there. I have been told this question involves complex numbers which I am not familiar with. Can someone please show me how to find the roots of this question with the quadratic formula?

Edit 2: The original question is to solve inequalities for $-x^4+3x^3-2x^2-16x+16<0$. I have found 2 roots already which are 1 and -2.

Answer 1

I am using the information that the underlying problem you are trying to solve is to find out when $–x^4+3x^3-2x^2-16x+16<0.$

This question does not really even make sense if $x$ could be a complex number, so it is reasonable to assume that the course where you got this question did not want you to use complex numbers. So let's use only the kinds of numbers you know about, which I suppose is the real numbers. (Even if you have not seen them called by that name, I will use it here, since it is much more convenient and precise to write "real numbers" rather than "the kind of numbers you have studied.")

I assume you have discovered that $-x^2+4x-8$ is a factor of $–x^4+3x^3-2x^2-16x+16.$

Now the fact that the quadratic equation asks you for the square root of a negative number is significant. It tells you that there cannot be any real number that is a root of $-x^2+4x-8$. In order for the quadratic equation to produce a real number, the square root would also have to be real (technically this is only because all the parts outside the square root are real, but they are). Now you may know that any real number, squared, produces a positive number (except zero, whose square is zero). And that is why you cannot find the square root of a negative number when you must look for the square root among the real numbers.

In fact, you can write $$-x^2+4x-8 = -(x^2 - 4x + 8) = -((x - 2)^2 + 4),$$ and now we see that for any real number $x,$ the quantity $x-2$ is also a number whose square is positive (or zero). When we add $4$ to that we always get a positive number. Then put the negative sign in front, and we always get a negative number. So $-x^2+4x-8$ is always negative.

You should have found two other factors of $–x^4+3x^3-2x^2-16x+16,$ so you have three factors altogether. In order for the product of all three factors to be negative, given that $-x^2+4x-8$ is always negative, you need to look for values of $x$ that either make both the other factors positive ($P\times P\times N \mapsto N$) or make both the other factors negative ($N\times N\times N \mapsto N$).

Answer 2

An equation $Ax^2+Bx+C=0$ with $A,B,C\in \Bbb R$ may or may not have any solution $x\in \Bbb R.$

For example $A=1,B=0,C=1:$ No $x\in \Bbb R$ satisfies $x^2+1=0.$

The Quadratic Formula says that if $A\ne 0$ then $$Ax^2+Bx+C=0\iff \left(x+\frac {B}{2A}\right)^2=\frac {B^2-4AC}{4A^2}.$$ So if $A, B,C\in \Bbb R$ with $A\ne 0$ and $B^2-4AC<0$ then there is no $x\in \Bbb R$ such that $Ax^2+Bx+C=0.$

Answer 3

Let's apply the quadratic formula step by step to see where you went wrong.

Recall that for $ax^2+bx+c=0$, $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$.

In your case, $-x^2+4x-8=0$.

$x=\frac{-4\pm\sqrt{(4^2)-((4)(-1)(-8))}}{2(-1)}$

$x=\frac{-4\pm\sqrt{16-32}}{-2}=2\pm\frac{\sqrt{-16}}{-2}=2\pm\frac{{4i}}{-2}=2\pm-2i$

Thus, our roots are not real and are $x=2+2i$ and $x=2-2i$.