$$ x^2-8x+c=0 $$
In the quadratic equation above, \(c\) is a constant. If the equation has two unique solutions, which of the following CANNOT be the value of \(c\)?
It is important to be able to master factoring quadratics so you complete these questions efficiently.
\(c=7\)
$$x^2-8x+7 =0$$
$$ (x-7)(x-1)=0 $$
$$ x=7 \text{ and } x=1 \checkmark $$
\(c=12\)
$$x^2-8x+12 =0 $$
$$ (x-6)(x-2)=0 $$
$$ x=6 \text{ and } x=2 \checkmark $$
\(c=15\)
$$x^2-8x+15 =0 $$
$$ (x-5)(x-3)=0 $$
$$ x=5 \text{ and } x=3 \checkmark$$
\(c=16\)
$$x^2-8x+16 =0$$
$$ (x-4)(x-4)=0 $$
$$ x=4 ✖ $$
We can inspect the discriminant of the quadratic (the portion under the square root of the quadratic formula):
$$ \text{Quadratic Formula} = \frac{-b\pm\sqrt{\colorbox{aqua}{$b^2-4ac$}}}{2a} $$
$$ \text{Discriminant} = b^2-4ac$$
The cases for the discriminant are as follows:
- \(D\gt0\): 2 real solutions
- \(D=0\): 1 real solutions
- \(D\lt0\): 0 real solutions
We want \(D \gt 0 \) so that the equation has 2 unique solutions.
$$ x^2-8x+c=0 $$
$$ b^2-4ac \gt 0 $$
$$ (-8)^2-4(1)(c) \gt 0 $$
$$ 64-4c \gt 0 $$
$$ -4c \gt -64 $$
$$ c \lt 16 $$
Three options satisfy this inequality, but \(\boxed{16}\) does not.