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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
\[ (1) \quad x^2 + 4 x -45 = 0 \]\[ (2) \quad 2 x^2 -8 x + 6 = 0 \]\[ (3) \quad x^2 + 15 x + 56 = 0 \]\[ (4) \quad -3 x^2 + 9 x + 12 = 0 \]\[ (5) \quad -2 x^2 + 3 x + 27 = 0 \]

Example (1)


\[ x^2 + 4 x -45 = 0 \]\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]\[ x = \frac{-(4) \pm \sqrt{(4)^2-4(1)(-45)}}{2(1)} \]\[ x = \frac{ -4 \pm \sqrt{ 196 }}{ 2 } \]\[ x = \frac{ -4 \pm 14}{ 2 } \]
\begin{align} x_1 & = \frac{ -4 - 14}{ 2 } \\ & = \frac{ -18}{ 2 } \\ & = -9 \end{align}\begin{align} x_2 & = \frac{ -4 + 14}{ 2 } \\ & = \frac{ 10}{ 2 } \\ & = 5 \end{align}
\[ \therefore x = -9 \quad \text{or} \quad 5 \]

Example (2)


\[ 2 x^2 -8 x + 6 = 0 \]\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]\[ x = \frac{-(-8) \pm \sqrt{(-8)^2-4(2)(6)}}{2(2)} \]\[ x = \frac{ 8 \pm \sqrt{ 16 }}{ 4 } \]\[ x = \frac{ 8 \pm 4}{ 4 } \]
\begin{align} x_1 & = \frac{ 8 - 4}{ 4 } \\ & = \frac{ 4}{ 4 } \\ & = 1 \end{align}\begin{align} x_2 & = \frac{ 8 + 4}{ 4 } \\ & = \frac{ 12}{ 4 } \\ & = 3 \end{align}
\[ \therefore x = 1 \quad \text{or} \quad 3 \]

Example (3)


\[ x^2 + 15 x + 56 = 0 \]\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]\[ x = \frac{-(15) \pm \sqrt{(15)^2-4(1)(56)}}{2(1)} \]\[ x = \frac{ -15 \pm \sqrt{ 1 }}{ 2 } \]\[ x = \frac{ -15 \pm 1}{ 2 } \]
\begin{align} x_1 & = \frac{ -15 - 1}{ 2 } \\ & = \frac{ -16}{ 2 } \\ & = -8 \end{align}\begin{align} x_2 & = \frac{ -15 + 1}{ 2 } \\ & = \frac{ -14}{ 2 } \\ & = -7 \end{align}
\[ \therefore x = -8 \quad \text{or} \quad -7 \]

Example (4)


\[ -3 x^2 + 9 x + 12 = 0 \]\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]\[ x = \frac{-(9) \pm \sqrt{(9)^2-4(-3)(12)}}{2(-3)} \]\[ x = \frac{ -9 \pm \sqrt{ 225 }}{ -6 } \]\[ x = \frac{ -9 \pm 15}{ -6 } \]
\begin{align} x_1 & = \frac{ -9 - 15}{ -6 } \\ & = \frac{ -24}{ -6 } \\ & = 4 \end{align}\begin{align} x_2 & = \frac{ -9 + 15}{ -6 } \\ & = \frac{ 6}{ -6 } \\ & = -1 \end{align}
\[ \therefore x = 4 \quad \text{or} \quad -1 \]

Example (5)


\[ -2 x^2 + 3 x + 27 = 0 \]\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]\[ x = \frac{-(3) \pm \sqrt{(3)^2-4(-2)(27)}}{2(-2)} \]\[ x = \frac{ -3 \pm \sqrt{ 225 }}{ -4 } \]\[ x = \frac{ -3 \pm 15}{ -4 } \]
\begin{align} x_1 & = \frac{ -3 - 15}{ -4 } \\ & = \frac{ -18}{ -4 } \\ & = \frac{ 9}{ 2 } \end{align}\begin{align} x_2 & = \frac{ -3 + 15}{ -4 } \\ & = \frac{ 12}{ -4 } \\ & = -3 \end{align}
\[ \therefore x = \frac{ 9}{ 2 } \quad \text{or} \quad -3 \]