18. Solving Quadratic Equations Using the Quadratic Formula

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Learning Objective:

Apply the quadratic formula to solve a quadratic equation in one variable when factoring is not possible or not easy.

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Solving Quadratic Equations Quadratic Formula Download

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The quadratic formula

We can always solve quadratic equations using the quadratic formula when factoring the quadratic equation either doesn’t work or isn’t particularly easy. The quadratic formula, gives solutions to the equation , ax²+bx+c = 0, where a, b, and c are real numbers and a ≠ 0.

If the discriminant b²− 4ac > 0, then there are two real solutions.
If b²− 4ac=0, there is one real solution.
If b²− 4ac < 0, there are no real solutions

Examples

Find any roots of x² + x – 6 = 0:

$x=\frac{-1\pm\sqrt{1^{2}-4(1)(-6)}}{2(1)}=\frac{-1\pm\sqrt{25}}{2}=\frac{-1\pm5}{2}=-3,2 \text{. Two real solutions.}$

$\text{Check: } (-3)^2 + (-3) - 6 = 9 - 3 - 6 = 0, \ 2^2 + 2 - 6 = 4 + 2 - 6 = 0.$

Find any roots of 4x² – 12x + 9 = 0:

$x=\frac{12\pm\sqrt{(-12)^{2}-4(4)(9)}}{2(4)}=\frac{12\pm\sqrt{0}}{8}=\frac{3}{2} \text{. One real solution.}$

$\text{Check: } 4\left(\frac{3}{2}\right)^2 - 12\left(\frac{3}{2}\right) + 9 = 9 - 18 + 9 = 0.$

Find any roots of x² + 4x + 5 = 0:

$x=\frac{-1\pm\sqrt{1^{2}-4(1)(5)}}{2(1)}=\frac{-1\pm\sqrt{-19}}{2} \text{. No real solution.}$

When factoring is not possible

Using the quadratic formula is most useful when factoring is not possible. For example:

Find any roots of 4x² − 8x + 1 = 0:

$x = \frac{8 \pm \sqrt{(-8)^2 - 4(4)(1)}}{2(4)} = \frac{8 \pm \sqrt{48}}{8} = \frac{8 \pm \sqrt{16(3)}}{8}$

$\quad = \frac{8 \pm 4\sqrt{3}}{8} = 1 \pm \frac{\sqrt{3}}{2} = 1 - \frac{\sqrt{3}}{2}, 1 + \frac{\sqrt{3}}{2} = 0.134, 1.866.$

$\text{Check: } 4(0.134)^2 - 8(0.134) + 1$

$= 0.072 - 1.072 + 1 = 0,$

$4(1.866)^2 - 8(1.866) + 1 = 13.928 - 14.928 + 1 = 0.$

When factoring isn’t easy

It is also useful when factoring isn’t particularly easy. For example:

Find any roots of 24x² + 22x − 35 = 0:

$x = \frac{-22 \pm \sqrt{22^2 - 4(24)(-35)}}{2(24)} = \frac{-22 \pm \sqrt{3844}}{48}$

$= \frac{-22 \pm 62}{48} = \frac{-11 \pm 31}{24} = -\frac{42}{24}, \frac{20}{24}$

$= -\frac{7}{4}, \frac{5}{6}.$

$\text{So, } 24x^2 + 22x - 35 = (4x+7)(6x-5).$

$\text{Check: } 24\left(-\frac{7}{4}\right)^2 + 22\left(-\frac{7}{4}\right) - 35 = \frac{147}{2} - \frac{77}{2} - \frac{70}{2} = 0,$

$24\left(\frac{5}{6}\right)^2 + 22\left(\frac{5}{6}\right) - 35 = \frac{50}{3} + \frac{55}{3} - \frac{105}{3} = 0.$

More examples

Find any roots of 2x² + 3x − 1 = 0:

$x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-1)}}{2(2)} = \frac{-3 \pm \sqrt{17}}{4} = -1.781, 0.281. \text{ Two solutions.}$

$\text{Check: } 2(-1.781)^2 + 3(-1.781) - 1 = 6.344 - 5.343 - 1 \approx 0,$

$\ 2(0.281)^2 + 3(0.281) - 1 = 0.158 + 0.843 - 1 \approx 0.$

Find any roots of 10x² − 19x + 6 = 0

$x = \frac{19 \pm \sqrt{(-19)^2 - 4(10)(6)}}{2(10)} = \frac{19 \pm \sqrt{121}}{20} = \frac{19 \pm 11}{20} = \frac{2}{5}, \frac{3}{2}. \text{ Two solutions.}$