Using the quadratic formula to solve quadratics (works every time)
The solutions of the quadratic equation $ax^2 + bx + c = 0$ are
$x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}.$
We call the quantity $\Delta = b^2-4ac$ (which we have already seen before!) the
discriminant of the quadratic ($\Delta$ is the Greek letter delta) and we have the following general principle:
- If $\Delta$ is positive, there are two distinct real solutions.
- If $\Delta$ is zero, there is only one real solution: $x = -\dfrac{b}{2a}.$ (Why?)
- If $\Delta$ is negative, there are no real solutions.
Examples
1. $x^2-5x-12 = 0$ has $a = 2, b = -5,$ #[and][y]# $c = -12.$ The discriminant is
$\Delta = b^2-4ac = (-5)^2-4(2)(-12) = 25 + 96 = 121,$
which is positive, so there are two real solutions:
$x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ \t ${}= \dfrac{5\pm\sqrt{(-5)^2-4(2)(-12)}}{2(2)}$
\\ \t ${}= \dfrac{5\pm\sqrt{121}}{4} = \dfrac{5\pm 11}{4}$
\\ \t ${}= \dfrac{16}{4}$ #[or][o]# $-\dfrac{6}{4}$
\\ \t ${}= 4$ #[or][o]# $-\dfrac{3}{2}$
2. $x^2+ 2x - 1 = 0$ has $a = 1, b = 2,$ #[and][y]# $c = -1.$ The discriminant is
$\Delta = b^2-4ac = 2^2-4(1)(-1) = 4+4 = 8,$
which is positive, so there are two real solutions:
$x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ \t ${}= \dfrac{-2\pm\sqrt{2^2-4(1)(-1)}}{2(1)}$
\\ \t ${}= \dfrac{-2\pm\sqrt{8}}{2} = \dfrac{-2\pm2\sqrt{2}}{2}$
\\ \t ${}= -1 + \sqrt{2}$ #[or][o]# $-1 - \sqrt{2}$
3. $4x^2 = 12x - 9$ can be rewritten as $4x^2-12x+9 = 0$ which has $a = 4, b = -12,$ #[and][y]# $c = 9.$The discriminant is
$\Delta = b^2-4ac = (-12)^2-4(4)(9) = 144-144 = 0,$
which is zero, so there is only one real solution:
$x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ \t ${}= \dfrac{12\pm\sqrt{(-12)^2-4(4)(9)}}{2(4)}$
\\ \t ${}= \dfrac{12\pm\sqrt{144-144}}{8} = \dfrac{12\pm\sqrt{0}}{8}= \dfrac{12}{8}$
\\ \t ${}= \dfrac{3}{2}$
4. $x^2+x+1 = 0$ has $a = 1, b = 1,$ #[and][y]# $c = 1.$ The discriminant is
$\Delta = b^2-4ac = 1^2-4(1)(1) = 1 - 4 = -3,$
which is negative, so there are no real solutions.