In many cases, factorising a quadratic equation or completing the square can be long or difficult. We can instead use the quadratic formula.
\( \begin{align} \displaystyle \require{AMSsymbols} \require{color}
ax^2 + bx + c &= 0 \\
ax^2 + bx &= -c \\
x^2 + \dfrac{b}{a}x &= -\dfrac{c}{a} \\
x^2 + \dfrac{b}{a}x \color{red} + \Big(\dfrac{b}{2a}\Big)^2 &= -\dfrac{c}{a} \color{red} + \Big(\dfrac{b}{2a}\Big)^2\\
\Big(x+\dfrac{b}{2a}\Big)^2 &= \dfrac{b^2-4ac}{4a^2} \\
x+\dfrac{b}{2a} &= \pm \sqrt{\dfrac{b^2-4ac}{4a^2}} \\
x &= -\dfrac{b}{2a} \pm \dfrac{\sqrt{b^2-4ac}}{2a} \\
\therefore x &= \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}
\end{align} \)
Example 1
Solve $x^2-2x-6 =0$ for $x$.
\( \begin{align} \displaystyle
a &= 1, b=-2,c=-6 \\
x &= \dfrac{-(-2) \pm \sqrt{(-2)^2-4 \times 1 \times (-6)}}{2 \times 1} \\
&= \dfrac{2 \pm \sqrt{4+24}}{2} \\
&= \dfrac{2 \pm \sqrt{28}}{2} \\
&= \dfrac{2 \pm 2\sqrt{7}}{2} \\
\therefore &= 1 \pm \sqrt{7}
\end{align} \)
Example 2
Solve $2x^2 + 3x-6 = 0$ for $x$.
\( \begin{align} \displaystyle
a &= 2, b=3,c=-6 \\
x &= \dfrac{-3 \pm \sqrt{3^2-4 \times 2 \times (-6)}}{2 \times 2} \\
&= \dfrac{-3 \pm \sqrt{9+48}}{4} \\
\therefore &= \dfrac{-3 \pm \sqrt{57}}{4}
\end{align} \)
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