E. Deriving the Quadratic Formula
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E. Deriving the Quadratic Formula
Rather than converting from the standard form of a quadratic equation into the vertex form by completing the square, and then solving for \(x\), you can derive, and then use a formula, called the
Quadratic Formula,
that utilizes the values \(a\), \(b\), and \(c\) to calculate the solutions to a quadratic equation.
What follows are the steps to derive the quadratic formula. To help clarify the process, write down the steps yourself on a separate sheet of paper. Also, watch the video showing the derivation of the formula.
Start by converting \(ax^2 + bx + c = 0\) from standard form into vertex form by completing the square.
Step 1: Group and factor out \(a\) from the first two terms.
Step 2: Add and subtract a constant, making a perfect square trinomial within the brackets.
Step 3: Remove the subtracted value within the brackets by multiplying by \(a\).
Step 4: Simplify the vertex form.
Next, solve for \(x\).
Step 5: Isolate the squared binomial.
Step 6: Take the square root of both sides.
Step 7: Isolate \(x\).
What follows are the steps to derive the quadratic formula. To help clarify the process, write down the steps yourself on a separate sheet of paper. Also, watch the video showing the derivation of the formula.
Start by converting \(ax^2 + bx + c = 0\) from standard form into vertex form by completing the square.
Step 1: Group and factor out \(a\) from the first two terms.
\[\begin{align}
ax^2 + bx + c &= 0 \\
a\left[ {x^2 + \left( {\frac{b}{a}} \right)x} \right] + c &= 0 \\
\end{align}\]
ax^2 + bx + c &= 0 \\
a\left[ {x^2 + \left( {\frac{b}{a}} \right)x} \right] + c &= 0 \\
\end{align}\]
Step 2: Add and subtract a constant, making a perfect square trinomial within the brackets.
\[\begin{align}
a\left[ {x^2 + \left( {\frac{b}{a}} \right)x} \right] + c &= 0 \\
a\left[ {x^2 + \left( {\frac{b}{a}} \right)x + \left( {\frac{b}{{2a}}} \right)^2 - \left( {\frac{b}{{2a}}} \right)^2 } \right] + c &= 0 \\
\end{align}\]
a\left[ {x^2 + \left( {\frac{b}{a}} \right)x} \right] + c &= 0 \\
a\left[ {x^2 + \left( {\frac{b}{a}} \right)x + \left( {\frac{b}{{2a}}} \right)^2 - \left( {\frac{b}{{2a}}} \right)^2 } \right] + c &= 0 \\
\end{align}\]
Step 3: Remove the subtracted value within the brackets by multiplying by \(a\).
\[\begin{align}
a\left[ {x^2 + \left( {\frac{b}{a}} \right)x + \left( {\frac{b}{{2a}}} \right)^2 - \left( {\frac{b}{{2a}}} \right)^2 } \right] + c &= 0 \\
a\left[ {x^2 + \left( {\frac{b}{a}} \right)x + \left( {\frac{b}{{2a}}} \right)^2 } \right] + c - \frac{{b^2 }}{{4a}} &= 0 \\
\end{align}\]
a\left[ {x^2 + \left( {\frac{b}{a}} \right)x + \left( {\frac{b}{{2a}}} \right)^2 - \left( {\frac{b}{{2a}}} \right)^2 } \right] + c &= 0 \\
a\left[ {x^2 + \left( {\frac{b}{a}} \right)x + \left( {\frac{b}{{2a}}} \right)^2 } \right] + c - \frac{{b^2 }}{{4a}} &= 0 \\
\end{align}\]
Step 4: Simplify the vertex form.
\[\begin{align}
a\left[ {x^2 + \left( {\frac{b}{a}} \right)x + \left( {\frac{b}{{2a}}} \right)^2 } \right] + c - \frac{{b^2 }}{{4a}} &= 0 \\
a\left( {x + \frac{b}{{2a}}} \right)^2 + \frac{{4ac - b^2 }}{{4a}} &= 0 \\
\end{align}\]
a\left[ {x^2 + \left( {\frac{b}{a}} \right)x + \left( {\frac{b}{{2a}}} \right)^2 } \right] + c - \frac{{b^2 }}{{4a}} &= 0 \\
a\left( {x + \frac{b}{{2a}}} \right)^2 + \frac{{4ac - b^2 }}{{4a}} &= 0 \\
\end{align}\]
Next, solve for \(x\).
Step 5: Isolate the squared binomial.
\[\begin{align}
a\left( {x + \frac{b}{{2a}}} \right)^2 + \frac{{4ac - b^2 }}{{4a}} &= 0 \\
a\left( {x + \frac{b}{{2a}}} \right)^2 &= \frac{{b^2 - 4ac}}{{4a}} \\
\left( {x + \frac{b}{{2a}}} \right)^2 &= \frac{{b^2 - 4ac}}{{4a^2 }} \\
\end{align}\]
a\left( {x + \frac{b}{{2a}}} \right)^2 + \frac{{4ac - b^2 }}{{4a}} &= 0 \\
a\left( {x + \frac{b}{{2a}}} \right)^2 &= \frac{{b^2 - 4ac}}{{4a}} \\
\left( {x + \frac{b}{{2a}}} \right)^2 &= \frac{{b^2 - 4ac}}{{4a^2 }} \\
\end{align}\]
Step 6: Take the square root of both sides.
\[\begin{align}
\left( {x + \frac{b}{{2a}}} \right)^2 &= \frac{{b^2 - 4ac}}{{4a^2 }} \\
\sqrt {\left( {x + \frac{b}{{2a}}} \right)^2 } &= \pm \sqrt {\frac{{b^2 - 4ac}}{{4a^2 }}} \\
x + \frac{b}{{2a}} &= \frac{{ \pm \sqrt {b^2 - 4ac} }}{{2a}} \\
\end{align}\]
\left( {x + \frac{b}{{2a}}} \right)^2 &= \frac{{b^2 - 4ac}}{{4a^2 }} \\
\sqrt {\left( {x + \frac{b}{{2a}}} \right)^2 } &= \pm \sqrt {\frac{{b^2 - 4ac}}{{4a^2 }}} \\
x + \frac{b}{{2a}} &= \frac{{ \pm \sqrt {b^2 - 4ac} }}{{2a}} \\
\end{align}\]
Step 7: Isolate \(x\).
\[\begin{align}
x + \frac{b}{{2a}} &= \frac{{ \pm \sqrt {b^2 - 4ac} }}{{2a}} \\
x &= - \frac{b}{{2a}} \pm \frac{{\sqrt {b^2 - 4ac} }}{{2a}} \\
x &= \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}} \\
\end{align}\]
x + \frac{b}{{2a}} &= \frac{{ \pm \sqrt {b^2 - 4ac} }}{{2a}} \\
x &= - \frac{b}{{2a}} \pm \frac{{\sqrt {b^2 - 4ac} }}{{2a}} \\
x &= \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}} \\
\end{align}\]
Key Lesson Marker |
Quadratic Formula
Quadratic Formula
Given the quadratic equation \(ax^2 + bx + c = 0\), where \(a\) does not equal \(0\), the quadratic formula
is
\[x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}, a \ne 0\]