Reduced formula for solving quadratic equations

When a quadratic equation has an even coefficient $$ ax^2+bx+c=0 $$ there is a convenient shortcut. In this case, we can use the reduced quadratic formula $$ x = \frac{-\frac{b}{2} \pm \sqrt{ \begin{pmatrix} \frac{b}{2} \end{pmatrix}^2 - ac }}{a}$$ which serves as an effective alternative to the standard version.

The reduced formula streamlines the computation because both the numerator and the denominator are divided by two, making the arithmetic cleaner.

A practical example

Consider the equation:

$$ 4x^2 - 4x - 3 = 0 $$

Here the coefficient b=-4 is even.

So we apply the reduced quadratic formula:

$$ x = \frac{-\frac{b}{2} \pm \sqrt{ \begin{pmatrix} \frac{b}{2} \end{pmatrix}^2 - ac }}{a}$$

Substituting a=4, b=-4 and c=-3 gives:

$$ x = \frac{-\frac{(-4)}{2} \pm \sqrt{ \begin{pmatrix} \frac{-4}{2} \end{pmatrix}^2 - (4)(-3) }}{4}$$

$$ x = \frac{2 \pm \sqrt{ (-2)^2 + 12 }}{4}$$

$$ x = \frac{2 \pm \sqrt{ 4 + 12 }}{4}$$

$$ x = \frac{2 \pm \sqrt{ 16 }}{4}$$

The discriminant is positive.

Therefore, the equation has two distinct real solutions:

$$ x = \frac{2 \pm 4}{4} = \begin{cases} x = \frac{2 - 4}{4} = - \frac{1}{2} \\ \\ x = \frac{2 + 4}{4} = \frac{3}{2} \end{cases} $$

Note. Had we used the standard quadratic formula the result would have been the same. The steps, however, would have involved slightly more cumbersome arithmetic. $$ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} $$ Substituting a=4, b=-4 and c=-3 we obtain $$ x = \frac{-(-4) \pm \sqrt{(-4)^2-4(4)(-3)}}{2(4)} $$ $$ x = \frac{4 \pm \sqrt{16+48}}{8} $$ $$ x = \frac{4 \pm \sqrt{64}}{8} $$ Since the discriminant is positive, the equation has two distinct real roots. $$ x = \frac{4 \pm 8}{8} \begin{cases} x = \frac{4 - 8}{8} = - \frac{1}{2} \\ \\ x = \frac{4 + 8}{4} = \frac{3}{2} \end{cases} $$

The derivation

The general quadratic formula is:

$$ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} $$

Factor 4 out of the expression under the square root:

$$ x = \frac{-b \pm \sqrt{4 \cdot (\frac{b^2}{4}-ac)}}{2a} $$

This allows the 4 to be taken outside the radical:

$$ x = \frac{-b \pm 2 \sqrt{\frac{b^2}{4}-ac}}{2a} $$

$$ x = \frac{-b \pm 2 \sqrt{ \begin{pmatrix} \frac{b}{2} \end{pmatrix}^2 -ac}}{2a} $$

Using the invariant property of fractions, we can divide both the numerator and denominator by two:

$$ x = \frac{ \frac{-b \pm 2 \sqrt{ \begin{pmatrix} \frac{b}{2} \end{pmatrix}^2 -ac} }{2} }{\frac{2a}{2}} $$

$$ x = \frac{\frac{-b}{2} \pm \frac{1}{2} \cdot 2 \sqrt{ \begin{pmatrix} \frac{b}{2} \end{pmatrix}^2 -ac} }{a} $$

$$ x = \frac{\frac{-b}{2} \pm \sqrt{ \begin{pmatrix} \frac{b}{2} \end{pmatrix}^2 -ac} }{a} $$

The final expression is the reduced form of the quadratic solution formula.

Note. In the reduced formula, the expression under the square root corresponds to the discriminant Δ of the standard formula divided by 4. $$ \frac{\Delta}{4} = \frac{b^2-4ac}{4} = \frac{b^2}{4} - ac = \begin{pmatrix} \frac{b}{2} \end{pmatrix}^2 -ac $$

And so on.