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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
\[ (1) \quad 5 x^2 + 11 x + 6 = 0 \]\[ (2) \quad 3 x^2 + 19 x + 28 = 0 \]\[ (3) \quad 2 x^2 + 10 x + 8 = 0 \]\[ (4) \quad 8 x^2 + 17 x + 9 = 0 \]\[ (5) \quad 3 x^2 + 11 x + 8 = 0 \]

Example (1)


\[ 5 x^2 + 11 x + 6 = 0 \]\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]\[ x = \frac{-(11) \pm \sqrt{(11)^2-4(5)(6)}}{2(5)} \]\[ x = \frac{ -11 \pm \sqrt{ 1 }}{ 10 } \]\[ x = \frac{ -11 \pm 1}{ 10 } \]
\begin{align} x_1 & = \frac{ -11 - 1}{ 10 } \\ & = \frac{ -12}{ 10 } \\ & = - \frac{ 6}{ 5 } \end{align}\begin{align} x_2 & = \frac{ -11 + 1}{ 10 } \\ & = \frac{ -10}{ 10 } \\ & = -1 \end{align}
\[ \therefore x = - \frac{ 6}{ 5 } \quad \text{or} \quad -1 \]

Example (2)


\[ 3 x^2 + 19 x + 28 = 0 \]\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]\[ x = \frac{-(19) \pm \sqrt{(19)^2-4(3)(28)}}{2(3)} \]\[ x = \frac{ -19 \pm \sqrt{ 25 }}{ 6 } \]\[ x = \frac{ -19 \pm 5}{ 6 } \]
\begin{align} x_1 & = \frac{ -19 - 5}{ 6 } \\ & = \frac{ -24}{ 6 } \\ & = -4 \end{align}\begin{align} x_2 & = \frac{ -19 + 5}{ 6 } \\ & = \frac{ -14}{ 6 } \\ & = - \frac{ 7}{ 3 } \end{align}
\[ \therefore x = -4 \quad \text{or} \quad - \frac{ 7}{ 3 } \]

Example (3)


\[ 2 x^2 + 10 x + 8 = 0 \]\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]\[ x = \frac{-(10) \pm \sqrt{(10)^2-4(2)(8)}}{2(2)} \]\[ x = \frac{ -10 \pm \sqrt{ 36 }}{ 4 } \]\[ x = \frac{ -10 \pm 6}{ 4 } \]
\begin{align} x_1 & = \frac{ -10 - 6}{ 4 } \\ & = \frac{ -16}{ 4 } \\ & = -4 \end{align}\begin{align} x_2 & = \frac{ -10 + 6}{ 4 } \\ & = \frac{ -4}{ 4 } \\ & = -1 \end{align}
\[ \therefore x = -4 \quad \text{or} \quad -1 \]

Example (4)


\[ 8 x^2 + 17 x + 9 = 0 \]\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]\[ x = \frac{-(17) \pm \sqrt{(17)^2-4(8)(9)}}{2(8)} \]\[ x = \frac{ -17 \pm \sqrt{ 1 }}{ 16 } \]\[ x = \frac{ -17 \pm 1}{ 16 } \]
\begin{align} x_1 & = \frac{ -17 - 1}{ 16 } \\ & = \frac{ -18}{ 16 } \\ & = - \frac{ 9}{ 8 } \end{align}\begin{align} x_2 & = \frac{ -17 + 1}{ 16 } \\ & = \frac{ -16}{ 16 } \\ & = -1 \end{align}
\[ \therefore x = - \frac{ 9}{ 8 } \quad \text{or} \quad -1 \]

Example (5)


\[ 3 x^2 + 11 x + 8 = 0 \]\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]\[ x = \frac{-(11) \pm \sqrt{(11)^2-4(3)(8)}}{2(3)} \]\[ x = \frac{ -11 \pm \sqrt{ 25 }}{ 6 } \]\[ x = \frac{ -11 \pm 5}{ 6 } \]
\begin{align} x_1 & = \frac{ -11 - 5}{ 6 } \\ & = \frac{ -16}{ 6 } \\ & = - \frac{ 8}{ 3 } \end{align}\begin{align} x_2 & = \frac{ -11 + 5}{ 6 } \\ & = \frac{ -6}{ 6 } \\ & = -1 \end{align}
\[ \therefore x = - \frac{ 8}{ 3 } \quad \text{or} \quad -1 \]