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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
\[ (1) \quad 7 x^2 -16 x + 9 = 0 \]\[ (2) \quad 9 x^2 + 20 x + 11 = 0 \]\[ (3) \quad 4 x^2 -2 x -6 = 0 \]\[ (4) \quad - x^2 -4 x + 12 = 0 \]\[ (5) \quad -2 x^2 -5 x + 3 = 0 \]

Example (1)


\[ 7 x^2 -16 x + 9 = 0 \]\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]\[ x = \frac{-(-16) \pm \sqrt{(-16)^2-4(7)(9)}}{2(7)} \]\[ x = \frac{ 16 \pm \sqrt{ 4 }}{ 14 } \]\[ x = \frac{ 16 \pm 2}{ 14 } \]
\begin{align} x_1 & = \frac{ 16 - 2}{ 14 } \\ & = \frac{ 14}{ 14 } \\ & = 1 \end{align}\begin{align} x_2 & = \frac{ 16 + 2}{ 14 } \\ & = \frac{ 18}{ 14 } \\ & = \frac{ 9}{ 7 } \end{align}
\[ \therefore x = 1 \quad \text{or} \quad \frac{ 9}{ 7 } \]

Example (2)


\[ 9 x^2 + 20 x + 11 = 0 \]\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]\[ x = \frac{-(20) \pm \sqrt{(20)^2-4(9)(11)}}{2(9)} \]\[ x = \frac{ -20 \pm \sqrt{ 4 }}{ 18 } \]\[ x = \frac{ -20 \pm 2}{ 18 } \]
\begin{align} x_1 & = \frac{ -20 - 2}{ 18 } \\ & = \frac{ -22}{ 18 } \\ & = - \frac{ 11}{ 9 } \end{align}\begin{align} x_2 & = \frac{ -20 + 2}{ 18 } \\ & = \frac{ -18}{ 18 } \\ & = -1 \end{align}
\[ \therefore x = - \frac{ 11}{ 9 } \quad \text{or} \quad -1 \]

Example (3)


\[ 4 x^2 -2 x -6 = 0 \]\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]\[ x = \frac{-(-2) \pm \sqrt{(-2)^2-4(4)(-6)}}{2(4)} \]\[ x = \frac{ 2 \pm \sqrt{ 100 }}{ 8 } \]\[ x = \frac{ 2 \pm 10}{ 8 } \]
\begin{align} x_1 & = \frac{ 2 - 10}{ 8 } \\ & = \frac{ -8}{ 8 } \\ & = -1 \end{align}\begin{align} x_2 & = \frac{ 2 + 10}{ 8 } \\ & = \frac{ 12}{ 8 } \\ & = \frac{ 3}{ 2 } \end{align}
\[ \therefore x = -1 \quad \text{or} \quad \frac{ 3}{ 2 } \]

Example (4)


\[ - x^2 -4 x + 12 = 0 \]\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]\[ x = \frac{-(-4) \pm \sqrt{(-4)^2-4(-1)(12)}}{2(-1)} \]\[ x = \frac{ 4 \pm \sqrt{ 64 }}{ -2 } \]\[ x = \frac{ 4 \pm 8}{ -2 } \]
\begin{align} x_1 & = \frac{ 4 - 8}{ -2 } \\ & = \frac{ -4}{ -2 } \\ & = 2 \end{align}\begin{align} x_2 & = \frac{ 4 + 8}{ -2 } \\ & = \frac{ 12}{ -2 } \\ & = -6 \end{align}
\[ \therefore x = 2 \quad \text{or} \quad -6 \]

Example (5)


\[ -2 x^2 -5 x + 3 = 0 \]\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]\[ x = \frac{-(-5) \pm \sqrt{(-5)^2-4(-2)(3)}}{2(-2)} \]\[ x = \frac{ 5 \pm \sqrt{ 49 }}{ -4 } \]\[ x = \frac{ 5 \pm 7}{ -4 } \]
\begin{align} x_1 & = \frac{ 5 - 7}{ -4 } \\ & = \frac{ -2}{ -4 } \\ & = \frac{ 1}{ 2 } \end{align}\begin{align} x_2 & = \frac{ 5 + 7}{ -4 } \\ & = \frac{ 12}{ -4 } \\ & = -3 \end{align}
\[ \therefore x = \frac{ 1}{ 2 } \quad \text{or} \quad -3 \]