Finding the zeros of a function simply means to set the domain equal to zero and then solve.
1. Let's do the first together:
\(x^2-4x+5=0\)
This equation is not factorable, so we have to use a different method other than factoring. Let's use the quadratic equation:
\(x = {-b\pm \sqrt{b^2-4ac} \over 2a}\)
a=1, b=-4, c=5
\(x = {-(-4) \pm \sqrt{(-4)^2-4(1)(5)} \over 2(1)}\)
Let's simplify this. First, I'll simplify what it is in the radical:
\(x = {4 \pm \sqrt{16-20} \over 2}\)
\(x = {4\pm2i\over 2}=2\pm i\)
The solution is 2+/- i.
2. Let's do the second one. Here is the original equation:
\(x^2+3x+4=0\)
This one, too, cannot be factored. I'll use the quadratic equation again:
\(x = {-3 \pm \sqrt{3^2-4(1)(4)} \over 2(1)}\)
I have simply substituted all the values into the equation. Now, solve for x:
\(x = {-3 \pm \sqrt{9-16} \over 2}\)
\(x = {-3 \pm \sqrt{-7} \over 2}\)
\(x = {-3\pm i\sqrt{7} \over2}\)
3. Here's the original equation:
\(-x^2+2x-3=0\)
I'll use the quadratic formula for the third time:
\(x = {-2 \pm \sqrt{2^2-4(-1)(-3)} \over 2(-1)}\)
\(x = {-2 \pm \sqrt{4-12} \over -2}\)
\(x = {-2\pm \sqrt{-8} \over -2}\)
\(x = {-2 \pm 2i\sqrt{2} \over -2}=1\pm i\sqrt{2}\)
4. Here's the original equation:
\(-x^2- 5x - 9=0\)
This is not factorable, so we must use the quadratic formula again.
\(x = {5\pm \sqrt{-11} \over-2}\)
\(x = {-5 \pm i\sqrt{11} \over2}\)
5. This one is the same story as the previous ones:
\(x = {1\pm\sqrt{-23} \over6}\)
\(x=\frac{1\pm i\sqrt{23}}{6}\)
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