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

Example (1)


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

Example (2)


\[ -2 x^2 + x + 6 = 0 \]\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]\[ x = \frac{-(1) \pm \sqrt{(1)^2-4(-2)(6)}}{2(-2)} \]\[ x = \frac{ -1 \pm \sqrt{ 49 }}{ -4 } \]\[ x = \frac{ -1 \pm 7}{ -4 } \]
\begin{align} x_1 & = \frac{ -1 - 7}{ -4 } \\ & = \frac{ -8}{ -4 } \\ & = 2 \end{align}\begin{align} x_2 & = \frac{ -1 + 7}{ -4 } \\ & = \frac{ 6}{ -4 } \\ & = - \frac{ 3}{ 2 } \end{align}
\[ \therefore x = 2 \quad \text{or} \quad - \frac{ 3}{ 2 } \]

Example (3)


\[ -2 x^2 - x + 6 = 0 \]\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]\[ x = \frac{-(-1) \pm \sqrt{(-1)^2-4(-2)(6)}}{2(-2)} \]\[ x = \frac{ 1 \pm \sqrt{ 49 }}{ -4 } \]\[ x = \frac{ 1 \pm 7}{ -4 } \]
\begin{align} x_1 & = \frac{ 1 - 7}{ -4 } \\ & = \frac{ -6}{ -4 } \\ & = \frac{ 3}{ 2 } \end{align}\begin{align} x_2 & = \frac{ 1 + 7}{ -4 } \\ & = \frac{ 8}{ -4 } \\ & = -2 \end{align}
\[ \therefore x = \frac{ 3}{ 2 } \quad \text{or} \quad -2 \]

Example (4)


\[ -3 x^2 + 8 x + 11 = 0 \]\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]\[ x = \frac{-(8) \pm \sqrt{(8)^2-4(-3)(11)}}{2(-3)} \]\[ x = \frac{ -8 \pm \sqrt{ 196 }}{ -6 } \]\[ x = \frac{ -8 \pm 14}{ -6 } \]
\begin{align} x_1 & = \frac{ -8 - 14}{ -6 } \\ & = \frac{ -22}{ -6 } \\ & = \frac{ 11}{ 3 } \end{align}\begin{align} x_2 & = \frac{ -8 + 14}{ -6 } \\ & = \frac{ 6}{ -6 } \\ & = -1 \end{align}
\[ \therefore x = \frac{ 11}{ 3 } \quad \text{or} \quad -1 \]

Example (5)


\[ 6 x^2 -8 x -8 = 0 \]\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]\[ x = \frac{-(-8) \pm \sqrt{(-8)^2-4(6)(-8)}}{2(6)} \]\[ x = \frac{ 8 \pm \sqrt{ 256 }}{ 12 } \]\[ x = \frac{ 8 \pm 16}{ 12 } \]
\begin{align} x_1 & = \frac{ 8 - 16}{ 12 } \\ & = \frac{ -8}{ 12 } \\ & = - \frac{ 2}{ 3 } \end{align}\begin{align} x_2 & = \frac{ 8 + 16}{ 12 } \\ & = \frac{ 24}{ 12 } \\ & = 2 \end{align}
\[ \therefore x = - \frac{ 2}{ 3 } \quad \text{or} \quad 2 \]