Solving second degree equations (deriving the quadratic formula)

The general form of a second degree equation is

x2+px+q=0(0)x^2 + px + q = 0 \tag{0}

where pp and qq can be any real number and xx is the roots of the equation. In equations of this form qq is the product of the roots and pp is the sum of the roots.

Equations of the form x2+q=0x^2 + q = 0 are easy to solve because p=0p = 0. The roots are simply ±xq±\sqrt{x} - q. The graphical representation of equations of this form is a curve that is symmetric about the xx axis. For instance the curve of y=x25y = x^2 - 5 looks like:

Now consider a more complicated equation like y=x26x5y = x^2 - 6x - 5. If we rewrite this equation as y=x(x6)5y = x(x - 6) - 5 we can see that y=5y = -5 when x(x6)=0x(x - 6) = 0. This could happen when x=0x = 0 or when x=6x = 6. The axis around which the curve is symmetric is half way between these values at the line x=3x = 3. We can confirm this by plotting the curve:

If we shifted the curve to the left by 3 units the axis of symmetry would be the xx axis. It is easy to find the roots of the equation of that line. We can do the shift algebraically by replacing xx with x+3x + 3 in the equation for yy.

In the general case, we are introducing a new equation whose roots are equal to the roots of the general second degree equation plus half the coeeficient of the xx term:

y=x+p2y = x + \frac{p}{2}

Therefore

x=yp2(1)x = y - \frac{p}{2} \tag{1}

Substituting this value of xx into the general equation gives:

(yp2)2+p(yp2)+q=0y2py2py2+p24+pyp22+q=0y2py+p24+pyp22+q=0y2p24+q=0\begin{aligned} (y - \frac{p}{2})^2 + p(y - \frac{p}{2}) + q &= 0 \\ y^2 - \frac{py}{2} - \frac{py}{2} + \frac{p^2}{4} + py - \frac{p^2}{2} + q &= 0 \\ y^2 - py + \frac{p^2}{4} +py - \frac{p^2}{2} + q &= 0 \\ y^2 - \frac{p^2}{4} + q &= 0 \end{aligned}

Therefore

y2=p24qy^2 = \frac{p^2}{4} - q

So our values of y are:

y=±p24qy = ±\sqrt{\frac{p^2}{4} - q}

Substituting this into equation (1)(1) gives:

x=p2±p24qx = -\frac{p}{2} ± \sqrt{\frac{p^2}{4} - q}

If we have an equation of the form

ax2+bx+c=0ax^2 + bx + c = 0

We can divide by aa to get

x2+(ba)x+ca=0x^2 + (\frac{b}{a})x + \frac{c}{a} = 0

This is the same form as equation (0)(0) above where p=bap = \frac{b}{a} and q=caq = \frac{c}{a}

Substituting these values into the equation (1)(1) gives

x=ba2±(ba)24ca=b2a±b24a2ca=b2a±b24ac4a2=b±b24ac2a(2)\begin{aligned} x &= -\frac{\frac{b}{a}}{2} ± \sqrt{\frac{(\frac{b}{a})^2}{4} - \frac{c}{a}} \\ &= -\frac{b}{2a} ± \sqrt{\frac{b^2}{4a^2} - \frac{c}{a}} \\ &= -\frac{b}{2a} ± \sqrt{\frac{b^2 - 4ac}{4a^2}} \\ &= -\frac{b ± \sqrt{b^2 - 4ac}}{2a} \tag{2} \end{aligned}

This is the quadratic formula.

Sources

Tags: Mathematics