Loading...

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
\[ (1) \quad 3 x^2 + 12 x + 9 = 0 \]\[ (2) \quad 5 x^2 + 13 x + 6 = 0 \]\[ (3) \quad x^2 + 3 x -108 = 0 \]\[ (4) \quad -5 x^2 + 2 x + 7 = 0 \]\[ (5) \quad - x^2 - x + 20 = 0 \]

Example (1)


\[ 3 x^2 + 12 x + 9 = 0 \]\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]\[ x = \frac{-(12) \pm \sqrt{(12)^2-4(3)(9)}}{2(3)} \]\[ x = \frac{ -12 \pm \sqrt{ 36 }}{ 6 } \]\[ x = \frac{ -12 \pm 6}{ 6 } \]
\begin{align} x_1 & = \frac{ -12 - 6}{ 6 } \\ & = \frac{ -18}{ 6 } \\ & = -3 \end{align}\begin{align} x_2 & = \frac{ -12 + 6}{ 6 } \\ & = \frac{ -6}{ 6 } \\ & = -1 \end{align}
\[ \therefore x = -3 \quad \text{or} \quad -1 \]

Example (2)


\[ 5 x^2 + 13 x + 6 = 0 \]\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]\[ x = \frac{-(13) \pm \sqrt{(13)^2-4(5)(6)}}{2(5)} \]\[ x = \frac{ -13 \pm \sqrt{ 49 }}{ 10 } \]\[ x = \frac{ -13 \pm 7}{ 10 } \]
\begin{align} x_1 & = \frac{ -13 - 7}{ 10 } \\ & = \frac{ -20}{ 10 } \\ & = -2 \end{align}\begin{align} x_2 & = \frac{ -13 + 7}{ 10 } \\ & = \frac{ -6}{ 10 } \\ & = - \frac{ 3}{ 5 } \end{align}
\[ \therefore x = -2 \quad \text{or} \quad - \frac{ 3}{ 5 } \]

Example (3)


\[ x^2 + 3 x -108 = 0 \]\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]\[ x = \frac{-(3) \pm \sqrt{(3)^2-4(1)(-108)}}{2(1)} \]\[ x = \frac{ -3 \pm \sqrt{ 441 }}{ 2 } \]\[ x = \frac{ -3 \pm 21}{ 2 } \]
\begin{align} x_1 & = \frac{ -3 - 21}{ 2 } \\ & = \frac{ -24}{ 2 } \\ & = -12 \end{align}\begin{align} x_2 & = \frac{ -3 + 21}{ 2 } \\ & = \frac{ 18}{ 2 } \\ & = 9 \end{align}
\[ \therefore x = -12 \quad \text{or} \quad 9 \]

Example (4)


\[ -5 x^2 + 2 x + 7 = 0 \]\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]\[ x = \frac{-(2) \pm \sqrt{(2)^2-4(-5)(7)}}{2(-5)} \]\[ x = \frac{ -2 \pm \sqrt{ 144 }}{ -10 } \]\[ x = \frac{ -2 \pm 12}{ -10 } \]
\begin{align} x_1 & = \frac{ -2 - 12}{ -10 } \\ & = \frac{ -14}{ -10 } \\ & = \frac{ 7}{ 5 } \end{align}\begin{align} x_2 & = \frac{ -2 + 12}{ -10 } \\ & = \frac{ 10}{ -10 } \\ & = -1 \end{align}
\[ \therefore x = \frac{ 7}{ 5 } \quad \text{or} \quad -1 \]

Example (5)


\[ - x^2 - x + 20 = 0 \]\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]\[ x = \frac{-(-1) \pm \sqrt{(-1)^2-4(-1)(20)}}{2(-1)} \]\[ x = \frac{ 1 \pm \sqrt{ 81 }}{ -2 } \]\[ x = \frac{ 1 \pm 9}{ -2 } \]
\begin{align} x_1 & = \frac{ 1 - 9}{ -2 } \\ & = \frac{ -8}{ -2 } \\ & = 4 \end{align}\begin{align} x_2 & = \frac{ 1 + 9}{ -2 } \\ & = \frac{ 10}{ -2 } \\ & = -5 \end{align}
\[ \therefore x = 4 \quad \text{or} \quad -5 \]