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Quadratics


The quadratic formula:
\(\frac{-b\pm \sqrt{b^2-4ac}}{2a}\)
If \( ax^2 + bx + c = 0 \), where \( a \ne 0 \), then: \[x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}\]

Solve the following equations using the quadratic formula
\[ (1) \quad x^2 -13 x + 36 = 0 \]\[ (2) \quad -2 x^2 -6 x + 20 = 0 \]\[ (3) \quad 3 x^2 + 11 x -4 = 0 \]\[ (4) \quad x^2 -17 x + 66 = 0 \]\[ (5) \quad 4 x^2 + 19 x + 22 = 0 \]

Example (1)


\[ x^2 -13 x + 36 = 0 \]\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]\[ x = \frac{-(-13) \pm \sqrt{(-13)^2-4(1)(36)}}{2(1)} \]\[ x = \frac{ 13 \pm \sqrt{ 25 }}{ 2 } \]\[ x = \frac{ 13 \pm 5}{ 2 } \]
\begin{align} x_1 & = \frac{ 13 - 5}{ 2 } \\ & = \frac{ 8}{ 2 } \\ & = 4 \end{align}\begin{align} x_2 & = \frac{ 13 + 5}{ 2 } \\ & = \frac{ 18}{ 2 } \\ & = 9 \end{align}
\[ \therefore x = 4 \quad \text{or} \quad 9 \]

Example (2)


\[ -2 x^2 -6 x + 20 = 0 \]\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]\[ x = \frac{-(-6) \pm \sqrt{(-6)^2-4(-2)(20)}}{2(-2)} \]\[ x = \frac{ 6 \pm \sqrt{ 196 }}{ -4 } \]\[ x = \frac{ 6 \pm 14}{ -4 } \]
\begin{align} x_1 & = \frac{ 6 - 14}{ -4 } \\ & = \frac{ -8}{ -4 } \\ & = 2 \end{align}\begin{align} x_2 & = \frac{ 6 + 14}{ -4 } \\ & = \frac{ 20}{ -4 } \\ & = -5 \end{align}
\[ \therefore x = 2 \quad \text{or} \quad -5 \]

Example (3)


\[ 3 x^2 + 11 x -4 = 0 \]\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]\[ x = \frac{-(11) \pm \sqrt{(11)^2-4(3)(-4)}}{2(3)} \]\[ x = \frac{ -11 \pm \sqrt{ 169 }}{ 6 } \]\[ x = \frac{ -11 \pm 13}{ 6 } \]
\begin{align} x_1 & = \frac{ -11 - 13}{ 6 } \\ & = \frac{ -24}{ 6 } \\ & = -4 \end{align}\begin{align} x_2 & = \frac{ -11 + 13}{ 6 } \\ & = \frac{ 2}{ 6 } \\ & = \frac{ 1}{ 3 } \end{align}
\[ \therefore x = -4 \quad \text{or} \quad \frac{ 1}{ 3 } \]

Example (4)


\[ x^2 -17 x + 66 = 0 \]\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]\[ x = \frac{-(-17) \pm \sqrt{(-17)^2-4(1)(66)}}{2(1)} \]\[ x = \frac{ 17 \pm \sqrt{ 25 }}{ 2 } \]\[ x = \frac{ 17 \pm 5}{ 2 } \]
\begin{align} x_1 & = \frac{ 17 - 5}{ 2 } \\ & = \frac{ 12}{ 2 } \\ & = 6 \end{align}\begin{align} x_2 & = \frac{ 17 + 5}{ 2 } \\ & = \frac{ 22}{ 2 } \\ & = 11 \end{align}
\[ \therefore x = 6 \quad \text{or} \quad 11 \]

Example (5)


\[ 4 x^2 + 19 x + 22 = 0 \]\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]\[ x = \frac{-(19) \pm \sqrt{(19)^2-4(4)(22)}}{2(4)} \]\[ x = \frac{ -19 \pm \sqrt{ 9 }}{ 8 } \]\[ x = \frac{ -19 \pm 3}{ 8 } \]
\begin{align} x_1 & = \frac{ -19 - 3}{ 8 } \\ & = \frac{ -22}{ 8 } \\ & = - \frac{ 11}{ 4 } \end{align}\begin{align} x_2 & = \frac{ -19 + 3}{ 8 } \\ & = \frac{ -16}{ 8 } \\ & = -2 \end{align}
\[ \therefore x = - \frac{ 11}{ 4 } \quad \text{or} \quad -2 \]