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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
\[ (1) \quad 3 x^2 + 24 x + 48 = 0 \]\[ (2) \quad 4 x^2 -14 x + 6 = 0 \]\[ (3) \quad x^2 -3 x + 2 = 0 \]\[ (4) \quad 3 x^2 + 4 x -7 = 0 \]\[ (5) \quad x^2 + 9 x + 14 = 0 \]

Example (1)


\[ 3 x^2 + 24 x + 48 = 0 \]\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]\[ x = \frac{-(24) \pm \sqrt{(24)^2-4(3)(48)}}{2(3)} \]\[ x = \frac{ -24 \pm \sqrt{ 0 }}{ 6 } \]\[ x = \frac{ -24 \pm 0}{ 6 } \]\begin{align} x & = \frac{ -24 }{ 6 } \\ & = -4 \end{align}\[ \therefore x = -4 \]

Example (2)


\[ 4 x^2 -14 x + 6 = 0 \]\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]\[ x = \frac{-(-14) \pm \sqrt{(-14)^2-4(4)(6)}}{2(4)} \]\[ x = \frac{ 14 \pm \sqrt{ 100 }}{ 8 } \]\[ x = \frac{ 14 \pm 10}{ 8 } \]
\begin{align} x_1 & = \frac{ 14 - 10}{ 8 } \\ & = \frac{ 4}{ 8 } \\ & = \frac{ 1}{ 2 } \end{align}\begin{align} x_2 & = \frac{ 14 + 10}{ 8 } \\ & = \frac{ 24}{ 8 } \\ & = 3 \end{align}
\[ \therefore x = \frac{ 1}{ 2 } \quad \text{or} \quad 3 \]

Example (3)


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

Example (4)


\[ 3 x^2 + 4 x -7 = 0 \]\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]\[ x = \frac{-(4) \pm \sqrt{(4)^2-4(3)(-7)}}{2(3)} \]\[ x = \frac{ -4 \pm \sqrt{ 100 }}{ 6 } \]\[ x = \frac{ -4 \pm 10}{ 6 } \]
\begin{align} x_1 & = \frac{ -4 - 10}{ 6 } \\ & = \frac{ -14}{ 6 } \\ & = - \frac{ 7}{ 3 } \end{align}\begin{align} x_2 & = \frac{ -4 + 10}{ 6 } \\ & = \frac{ 6}{ 6 } \\ & = 1 \end{align}
\[ \therefore x = - \frac{ 7}{ 3 } \quad \text{or} \quad 1 \]

Example (5)


\[ x^2 + 9 x + 14 = 0 \]\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]\[ x = \frac{-(9) \pm \sqrt{(9)^2-4(1)(14)}}{2(1)} \]\[ x = \frac{ -9 \pm \sqrt{ 25 }}{ 2 } \]\[ x = \frac{ -9 \pm 5}{ 2 } \]
\begin{align} x_1 & = \frac{ -9 - 5}{ 2 } \\ & = \frac{ -14}{ 2 } \\ & = -7 \end{align}\begin{align} x_2 & = \frac{ -9 + 5}{ 2 } \\ & = \frac{ -4}{ 2 } \\ & = -2 \end{align}
\[ \therefore x = -7 \quad \text{or} \quad -2 \]