The quadratic formula

Quadratic equations: Solving quadratic equations

The quadratic formula

The equation #x^2+5x+5=0# is an example of a quadratic equation that cannot be solved by factorizing. We could solve this equation by completing the square, but this method can be quite time-consuming, especially as equations become more complex.

It is for this reason that the quadratic formula, which is derived from the method of completing the square, is commonly used to solve quadratic equations which cannot be solved by factorizing.

Quadratic formula and discriminant

The solutions of a quadratic equation in the form \[\blue ax^2+\green bx+\purple c=0\] are:

\[x=\frac{-\green b-\sqrt{\orange D}}{2 \blue a} \lor x=\frac{-\green b+\sqrt{\orange D}}{2 \blue a}\]

#\orange D# is called the discriminant and is calculated as follows:

\[\orange D =\green b^2-4\blue a \purple c\]

The value of #\orange D# determines how many solutions there are:

If #\orange D \lt 0# then there are no solutions.

If #\orange D = 0# there is one solution, namely #x=\frac{-\green b}{2\blue a}#.

If #\orange D \gt 0# then there are two solutions.

We call this the quadratic formula.

geogebra picture

\[\begin{array}{rcl}\blue a x^2+\green b x +\purple c&=&0 \\
&&\phantom{xx}\blue{\text{the original equation}}\\
x^2+\frac{\green b}{\blue a} x +\frac{\purple c}{\blue a}&=&0 \\
&&\phantom{xx}\blue{\text{divided both sides by }a } \\
\left(x+\frac{\green b}{2\blue a}\right)^2-\left(\frac{\green b}{2 \blue a}\right)^2+\frac{\purple c}{\blue a}&=&0 \\ &&\phantom{xx}\blue{\text{completed the square}}\\ \left(x+\frac{\green b}{2\blue a}\right)^2-\frac{\green b^2}{4 \blue a^2}+\frac{4 \blue a\purple c}{4 \blue a^2}&=&0 \\ &&\phantom{xx}\blue{\text{simplified }\left(\frac{b}{2a}\right)^2\text{ and created a common denominator}}\\
\left(x+\frac{\green b}{2\blue a}\right)^2-\frac{\green b^2-4 \blue a\purple c}{4 \blue a^2}&=&0\\ &&\phantom{xx}\blue{\text{added the fractions}}\\
\left(x+\frac{\green b}{2\blue a}\right)^2&=&\frac{\green b^2-4 \blue a\purple c}{4 \blue a^2}\\
&&\phantom{xx}\blue{\text{transferred the constant to the right-hand side}}\\
x+\frac{\green b}{2\blue a}=-\sqrt{\frac{\green b^2-4 \blue a\purple c}{4 \blue a^2}} &\lor& x+\frac{\green b}{2\blue a}=\sqrt{\frac{\green b^2-4 \blue a\purple c}{4 \blue a^2}} \\
&&\phantom{xx}\blue{\text{took the square root of both sides}}\\ x=\frac{-\green b}{2\blue a}-\sqrt{\frac{\green b^2-4 \blue a\purple c}{4 \blue a^2}} &\lor& x=\frac{-\green b}{2\blue a}+\sqrt{\frac{\green b^2-4 \blue a\purple c}{4 \blue a^2}} \\&&\phantom{xx}\blue{\text{subtracted }\frac{b}{2a} \text{ from both sides}}\\
x=\frac{-\green b}{2\blue a}-\frac{\sqrt{\green b^2-4 \blue a\purple c}}{\sqrt{4 \blue a^2}} &\lor& x=\frac{-\green b}{2\blue a}+\frac{\sqrt{\green b^2-4 \blue a\purple c}}{\sqrt{4 \blue a^2}} \\&&\phantom{xx}\blue{\text{wrote the root of a fraction as the fraction of two roots}} \\
x=\frac{-\green b}{2\blue a}-\frac{\sqrt{\green b^2-4 \blue a\purple c}}{2\blue a} &\lor& x=\frac{-\green b}{2\blue a}+\frac{\sqrt{\green b^2-4 \blue a\purple c}}{2 \blue a} \\&&\phantom{xx}\blue{\text{simplified }} \\
x=\frac{-\green b-\sqrt{\green b^2-4\blue a \purple c}}{2\blue a} &\lor& x=\frac{-\green b+\sqrt{\green b^2-4\blue a \purple c}}{2\blue a} \\
&&\phantom{xx}\blue{\text{added the fractions}}\end{array}\]

Writing the expression #\green b^2-4\blue a \purple c# as #\orange D# results in the quadratic formula.

The quadratic formula can be used to solve any quadratic equation.

Solving quadratic equations using the quadratic formula

	Procedure	example
	Solving a quadratic equation for #x# using the quadratic formula	#3x^2+2x+5=4x+8#
Step 1	Reduce the equation to the form #\blue ax^2+\green bx+\purple c=0#.	#\blue 3x^2\green{-2}x\purple{-3}=0#
Step 2	Identify #\blue a#, #\green b# and #\purple c#.	#\blue a =\blue3#, #\green b=\green{-2}# and #\purple c = \purple{-3}#
Step 3	Calculate the discriminant #\orange D=\green b^2-4\blue a \purple c#.	#\orange D=(\green{-2})^2-4\cdot \blue3 \cdot (\purple{-3})=\orange{40}\gt0#
Step 4	Determine the number of solutions. If #D \gt 0#, there are #2# solutions. If #D=0#, there is one solution. If #D \lt 0#, then there are no solutions.	There are two solutions.
Step 5	Calculate the solutions, simplifying as much as possible. \[x=\frac{-\green b-\sqrt{\orange D}}{2 \blue a} \lor x=\frac{-\green b+\sqrt{\orange D}}{2 \blue a}\]	# \begin{array}{rcl}x&=&\dfrac{-(\green{-2})-\sqrt{\orange{40}}}{2 \cdot \blue{3}}=\dfrac{1}{3}-\dfrac{\sqrt{10}}{3} \\ &\lor& \\ x&=&\dfrac{-(\green{-2})+\sqrt{\orange{40}}}{2 \cdot \blue3}=\dfrac{1}{3}+\dfrac{\sqrt{10}}{3}\end {array} #

Solve #x# in\[x^2+x-1=0\]
Write:

#none# if there is no solution,
#x=a# if there is one solution,
#x=a \lor x=b# if there are two solutions,

with the correct exact values (no decimal numbers) of #a# and #b#.

#x={{-\sqrt{5}-1}\over{2}}\lor x={{\sqrt{5}-1}\over{2}}#

We solve the equation with the quadratic formula. The equation is already reduced to #0#. Hence, we start at step 2 of the procedure.

Step 2	We identify #a#, #b# and #c#. #a=1#, #b=1# and #c=-1#
Step 3	We calculate the discriminant. \[\begin{array}{rcl}D&=&b^2-4ac \\&&\phantom{xxx}\blue{\text{formula for discriminant}}\\ &=& 1^2-4\cdot 1\cdot (-1)\\&&\phantom{xxx}\blue{\text{formula entered}}\\ &=& 5 \\&&\phantom{xxx}\blue{\text{calculated}}\end{array}\]
Step 4	The discriminant #D \gt 0#, hence, the number of solutions is #2#.
Step 5	We determine the solutions of the equation. \[\begin{array}{rcl}x=\dfrac{-b-\sqrt{D}}{2a} &\lor& x=\dfrac{-b+\sqrt{D}}{2a} \\&&\phantom{xxx}\blue{\text{formula for solutions}}\\ x=\dfrac{{-1}-\sqrt{5}}{2 \cdot 1} &\lor& x=\dfrac{{-1}+\sqrt{5}}{2 \cdot 1} \\ &&\phantom{xxx}\blue{\text{formula entered }}\\ x= \displaystyle {{-\sqrt{5}-1}\over{2}} &\lor& x= \displaystyle {{\sqrt{5}-1}\over{2}} \\ &&\phantom{xxx}\blue{\text{simplified}}\\ \end{array}\]

The graph of the formula #y=x^2+x-1# is drawn below. The #x#-coordinates of the intersection points of the graph with the #x#-axis are the solutions of the equation.

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