10.5 Solving quadratic equations using the quadratic formula

In the equation $3x^2+5x+2=0,$

$\begin{array}{lll}a\hfill & =\hfill & 3\hfill \\ b\hfill & =\hfill & 5\hfill \\ c\hfill & =\hfill & 2\hfill \end{array}$

In the equation $12x^2-2x-1=0,$

$\begin{array}{lll}a\hfill & =\hfill & 12\hfill \\ b\hfill & =\hfill & -2\hfill \\ c\hfill & =\hfill & -1\hfill \end{array}$

In the equation $2y^2+3=0,$

$\begin{array}{lllll}a\hfill & =\hfill & 2\hfill & \hfill & \hfill \\ b\hfill & =\hfill & 0\hfill & \hfill & \text{Because\hspace{0.17em}the\hspace{0.17em}equation\hspace{0.17em}could\hspace{0.17em}be\hspace{0.17em}written}\hfill \\ \hfill & \hfill & \hfill & \hfill & 2y^2+0y+3=0\hfill \\ c\hfill & =\hfill & 3\hfill & \hfill & \hfill \end{array}$

In the equation $-8y^2+11y=0,$

$\begin{array}{lllll}a\hfill & =\hfill & -8\hfill & \hfill & \hfill \\ b\hfill & =\hfill & 11\hfill & \hfill & \hfill & \hfill \\ c\hfill & =\hfill & 0\hfill & \text{Since\hspace{0.17em}}-{\text{8y}}^{\text{2}}+11y+0=0.\hfill \end{array}$

In the equation $z^2=z+8,$

$\begin{array}{lllll}a\hfill & =\hfill & 1\hfill & \hfill & \hfill \\ b\hfill & =\hfill & -1\hfill & \hfill & \hfill \\ c\hfill & =\hfill & -8\hfill & \hfill & \text{When\hspace{0.17em}we\hspace{0.17em}write\hspace{0.17em}the\hspace{0.17em}equation\hspace{0.17em}in\hspace{0.17em}standard\hspace{0.17em}form,\hspace{0.17em}we\hspace{0.17em}get\hspace{0.17em}}{z}^2-z-8=0\hfill \end{array}$

Subtract $c$ from both sides.

$a{x}^2+bx=-c$

Divide both sides by $a,$ the coefficient of $x^2.$

$x^2+\frac{b}{a}x=\frac{-c}{a}$

Now we have the proper form to complete the square. Take one half the coefficient of $x,$ square it, and add the result to both sides of the equation found in step 2.

(a) $\frac{1}{2}·\frac{b}{a}=\frac{b}{2a}$ is one half the coefficient of $x.$

(b) ${\left(\frac{b}{2a}\right)}^2$ is the square of one half the coefficient of $x.$

$x^2+\frac{b}{a}x+{\left(\frac{b}{2a}\right)}^2=\frac{-c}{a}+{\left(\frac{b}{2a}\right)}^2$

The left side of the equation is now a perfect square trinomial and can be factored. This gives us

${\left(x+\frac{b}{2a}\right)}^2=\frac{-c}{a}+\frac{b^2}{4a^2}$

Add the two fractions on the right side of the equation. The LCD $=4a^2.$

$\begin{array}{lll}{\left(x+\frac{b}{2a}\right)}^2\hfill & =\hfill & \frac{-4ac}{4a^2}+\frac{b^2}{4a^2}\hfill \\ {\left(x+\frac{b}{2a}\right)}^2\hfill & =\hfill & \frac{-4ac+b^2}{4a^2}\hfill \\ {\left(x+\frac{b}{2a}\right)}^2\hfill & =\hfill & \frac{b^2-4ac}{4a^2}\hfill \end{array}$

Solve for $x$ using the method of extraction of roots.

$\begin{array}{lllll}x+\frac{b}{2a}\hfill & =\hfill & \pm \sqrt{\frac{b^2-4ac}{4a^2}}\hfill & \hfill & \hfill \\ x+\frac{b}{2a}\hfill & =\hfill & \pm \frac{\sqrt{b^2-4ac}}{\sqrt{4a^2}}\hfill & \hfill & \sqrt{4a^2}=\left|2a\right|=2\left|a\right|=\pm 2a\hfill \\ x+\frac{b}{2a}\hfill & =\hfill & \pm \frac{\sqrt{b^2-4ac}}{2a}\hfill & \hfill & \hfill \\ \hfill x& =\hfill & -\frac{b}{2a}\pm \frac{\sqrt{b^2-4ac}}{2a}\hfill & \hfill & \text{Add\hspace{0.17em}these\hspace{0.17em}two\hspace{0.17em}fractions}\text{.}\hfill \\ \hfill x& =\hfill & \frac{-b\pm \sqrt{b^2-4ac}}{2a}\hfill & \hfill & \hfill \end{array}$

$3x^2+5x+2=0.$

1. Identify $a,b,$ and $c.$
   
   $\begin{array}{llllll}a=3,\hfill & b=5,\hfill & \hfill & \text{and}\hfill & \hfill & c=2\hfill \end{array}$
2. Write the quadratic formula.
   
   $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
3. Substitute.
   
   $\begin{array}{lllll}x\hfill & =\hfill & \frac{-5\pm \sqrt{{\left(5\right)}^2-4\left(3\right)\left(2\right)}}{2\left(3\right)}\hfill & \hfill & \hfill \\ \hfill & =\hfill & \frac{-5\pm \sqrt{25-24}}{6}\hfill & \hfill & \hfill \\ \hfill & =\hfill & \frac{-5\pm \sqrt{1}}{6}\hfill & \hfill & \hfill \\ \hfill & =\hfill & \frac{-5\pm 1}{6}\hfill & \hfill & -5+1=-4\text{\hspace{0.17em}and\hspace{0.17em}}-5-1=-6\hfill \\ \hfill & =\hfill & \frac{-4}{6},\frac{-6}{6}\hfill & \hfill & \hfill \\ x\hfill & =\hfill & \frac{-2}{3},-1\hfill & \hfill & \hfill \end{array}$
   $\begin{array}{lll}Note:\hfill & \hfill & \text{Since\hspace{0.17em}these\hspace{0.17em}roots\hspace{0.17em}are\hspace{0.17em}rational\hspace{0.17em}numbers,\hspace{0.17em}this\hspace{0.17em}equation\hspace{0.17em}could\hspace{0.17em}have\hspace{0.17em}been\hspace{0.17em}solved\hspace{0.17em}by\hspace{0.17em}factoring}\text{.}\hfill \end{array}$

$12x^2-2x-1=0.$

1. Identify $a,b,$ and $c.$
   
   $\begin{array}{llllll}a=12,\hfill & b=-2,\hfill & \hfill & \text{and}\hfill & \hfill & c=-1\hfill \end{array}$
2. Write the quadratic formula.
   
   $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
3. Substitute.
   
   $$\begin{array}{lllll}x\hfill & =\hfill & \frac{-\left(-2\right)\pm \sqrt{{\left(-2\right)}^2-4\left(12\right)\left(-1\right)}}{2\left(12\right)}\hfill & \hfill & \hfill \\ \hfill & =\hfill & \frac{2\pm \sqrt{4+48}}{24}\hfill & \hfill & \text{Simplify}.\hfill \\ \hfill & =\hfill & \frac{2\pm \sqrt{52}}{24}\hfill & \hfill & \text{Simplify}.\hfill \\ \hfill & =\hfill & \frac{2\pm \sqrt{4·13}}{24}\hfill & \hfill & \text{Simplify}.\hfill \\ \hfill & =\hfill & \frac{2\pm 2\sqrt{13}}{24}\hfill & \hfill & \text{Reduce}.\text{\hspace{0.17em}Factor\hspace{0.17em}2\hspace{0.17em}from\hspace{0.17em}the\hspace{0.17em}terms\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}numerator}.\hfill \\ \hfill & =\hfill & \frac{2\left(1\pm \sqrt{13}\right)}{24}\hfill & \hfill & \hfill \\ x\hfill & =\hfill & \frac{1\pm \sqrt{13}}{12}\hfill & \hfill & \hfill \end{array}$$

$2y^2+3=0$

1. Identify $a,b,$ and $c.$
   
   $\begin{array}{llllll}a=2,\hfill & b=0,\hfill & \hfill & \text{and}\hfill & \hfill & c=3\hfill \end{array}$
2. Write the quadratic formula.
   
   $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
3. Substitute.
   
   $\begin{array}{lll}x\hfill & =\hfill & \frac{-0\pm \sqrt{0^2-4\left(2\right)\left(3\right)}}{2\left(2\right)}\hfill \\ x\hfill & =\hfill & \frac{0\pm \sqrt{-24}}{4}\hfill \end{array}$
   
   This equation has no real number solution since we have obtained a negative number under the radical sign.

$-8x^2+11x=0$

1. Identify $a,b,$ and $c.$
   
   $\begin{array}{llllll}a=-8,\hfill & b=11,\hfill & \hfill & \text{and}\hfill & \hfill & c=0\hfill \end{array}$
2. Write the quadratic formula.
   
   $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
3. Substitute.
   
   $\begin{array}{lllll}x\hfill & =\hfill & \frac{-11\pm \sqrt{{11}^2-4\left(-8\right)\left(0\right)}}{2\left(-8\right)}\hfill & \hfill & \hfill \\ \hfill & =\hfill & \frac{-11\pm \sqrt{121-0}}{-16}\hfill & \hfill & \text{Simplify}.\hfill \\ \hfill & =\hfill & \frac{-11\pm \sqrt{121}}{-16}\hfill & \hfill & \text{Simplify}.\hfill \\ \hfill & =\hfill & \frac{-11\pm 11}{-16}\hfill & \hfill & \hfill \\ x\hfill & =\hfill & 0,\frac{11}{8}\hfill & \hfill & \hfill \end{array}$

$\left(3x+1\right)\left(x-4\right)=x^2+x-2$

1. Write the equation in standard form.
   
   $\begin{array}{lll}3x^2-11x-4\hfill & =\hfill & x^2+x-2\hfill \\ 2x^2-12x-2\hfill & =\hfill & 0\hfill \\ x^2-6x-1\hfill & =\hfill & 0\hfill \end{array}$
2. Identify $a,b,$ and $c.$
   
   $\begin{array}{llllll}a=1,\hfill & b=-6,\hfill & \hfill & \text{and}\hfill & \hfill & c=-1\hfill \end{array}$
3. Write the quadratic formula.
   
   $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
4. Substitute.
   
   $\begin{array}{lll}x\hfill & =\hfill & \frac{-\left(-6\right)\pm \sqrt{{\left(-6\right)}^2-4\left(1\right)\left(-1\right)}}{2\left(1\right)}\hfill \\ \hfill & =\hfill & \frac{6\pm \sqrt{36+4}}{2}\hfill \\ \hfill & =\hfill & \frac{6\pm \sqrt{40}}{2}\hfill \\ \hfill & =\hfill & \frac{6\pm \sqrt{4·10}}{2}\hfill \\ \hfill & =\hfill & \frac{6\pm 2\sqrt{10}}{2}\hfill \\ \hfill & =\hfill & \frac{2\left(3\pm \sqrt{10}\right)}{2}\hfill \\ x\hfill & =\hfill & 3\pm \sqrt{10}\hfill \end{array}$