A standard form for a quadratic equation is
\(ax^2+bx+c=0 \quad , a \neq 0\)
where \(a\), \(b\) and \(c\) are constants.
The quadratic formula is
\(x = \dfrac{- b\pm\sqrt{b^2-4a c}}{2a}\)
* Notice that \(x\) has 2 values! So we get either
\(x = \dfrac{- b{\color{red}{+}}\sqrt{b^2-4a c}}{2a} \quad \)
or \(\quad x = \dfrac{- b{\color{red}{-}}\sqrt{b^2-4a c}}{2a} \)
Special Cases
1 Case 1
Normally a quadratic formula has 2 values. However, if \(b^2-4ac =0\) , \(x\) has only 1 value.
\( x = \dfrac{- b\pm\sqrt{b^2-4ac}}{2a} \\ \, \\ \quad \downarrow \\ \, \\ x = \dfrac{-b\pm\sqrt{0}}{2a} \\ \, \\ \quad \downarrow \\ \, \\ x = \dfrac{-b}{2a} \)
2 Case 2
If \( b^2-4ac \) < 0, \(x\) has no value.
\( \\ \, \\ x = \dfrac{- b\pm\sqrt{b^2-4ac}}{2a} \)
\( \quad \downarrow \\ \, \\ x = \dfrac{- b\pm\sqrt{\text{negative}}}{2a} \\ \, \\ \quad \downarrow \)
No value!
Example Solve \(2x^2+4x+1\).
Solution Using the quadratic formula, we can see that
\(a=2\), \(b=4\), \(c=1\)
\( \require{cancel} x = \dfrac{- b\pm\sqrt{b^2-4a c}}{2a} \\ \, \\ x = \dfrac{- 4\pm\sqrt{4^2-4(2)(1)}}{2(2)} \\ \, \\ x = \dfrac{- 4\pm\sqrt{16-8}}{4} \\ \, \\ x = \dfrac{- 4\pm\sqrt{8}}{4} \\ \, \\ x = \dfrac{- \cancel{{\color{red}{4}}}\pm\sqrt{8}}{\cancel{{\color{red}{4}}}} \\ \, \\ x = \pm \sqrt{8} \\ \, \\ x = \pm \sqrt{4 \times 2} \\ \, \\ x = \pm 2\sqrt{2} \\ \, \\ \)
So we have either \(x= 2\sqrt{2} \)
or \(x= -2\sqrt{2} \).
Solve \(x^2+5x-2=0\) using the quadratic formula.