Quadratic equations

Basic algebra skills: Quadratic functions

Quadratic equations

In addition to linear equations, there are quadratic equations. We will first look at what a quadratic equation is, and then we will look at a method of solving special cases of quadratic equations.

Quadratic equations

An equation with unknown #x# is called quadratic or second degree if the only terms appearing in it are constants, constant multiples of #x#, and constant multiples of #x^2#.

The general form is: \[a \cdot x^2+b \cdot x +c=0\]

Here #a#, #b# and #c# are numbers.

As with linear equations, we can reduce quadratic equations by applying the same operations to the left- and right-hand sides of the equation, such as addition and subtraction. This way, we can reduce any quadratic equation to its standard form, that is, the general form given in the definition. Starting from this standard form, we can use a number of solution methods.

Examples

Here are some examples of quadratic equations

\[\begin{array}{rcl} 2x^2-5x+7&=&x^2-2x+3\\ x^2 &=&0\\ x^2+1&=&0\end{array}\]

The first equation can be reduced to the general form #x^2-3x-4=0# by moving #x^2-2x+3# to the left-hand side. Substituting #x=-1# and #x=4# gives #0=0#; so these are two solutions. We will see later that these are the only solutions.

The second equation has exactly one solution: #x=0#.

The third equation has no solution, because #x^2+1\gt0# for all real numbers #x#.

In general, a quadratic equation has zero, one, or two solutions.

The left-hand side of the equation #a\cdot x^2+b\cdot x + c = 0# is a quadratic function. The equation can therefore be seen as the search for the point #x# in which the value of this quadratic function is zero; that is, the point where the graph of this function intersects the #x# axis.

We will discuss the general solution of a quadratic equation later. Now, we discuss the special case in which we can factorise the left-hand side of the equation. By doing this, the solution can quickly be found:

Solution of a Quadratic Equation by Factoring Consider the quadratic equation with unknown #x#: \[a\cdot x^2+b\cdot x+c=0\tiny,\] where #a#, #b# and #c# are real numbers with #a\ne0#.

If the left-hand side can be factorised into linear factors: \[ a\cdot x^2+b\cdot x+c=(x-p)\cdot(x-q)\tiny,\] where #p# and #q# are real numbers, then the solution to the equation is #x=p\lor x=q#.

The following rule has been used below

#A\cdot B=0\phantom{xxx}# is equivalent to #\phantom{xxx}A=0\lor B=0#, where #A = x-p# and #B = x-q# are:

\[\begin{array}{rcl}a\cdot x^2+b\cdot x+c&=&0\\ &&\phantom{xx}\color{blue}{\text{original equation}}\\ (x-p)\cdot(x-q)&=&0\\&&\phantom{xx}\color{blue}{\text{replaced left-hand side by factoring}}\\ x-p=0&\lor& x-q=0\\&&\phantom{xx}\color{blue}{\text{above mentioned rule}}\\ x=p&\lor& x=q \\&&\phantom{xx}\color{blue}{\text{constants to right-hand side}}\\ \end{array}\]

Approach

To factor the quadratic equation, we must first reduce it to the form:

\[x^2 + b \cdot x+c=0\]

We do this by first reducing it to the standard form (we call this reduction to #0#) and then dividing all terms by the coefficient of #x^2#.

For example: #3x^2-24=-6x# can be reduced to the standard form #3x^2+6x-24=0#. If we then divide all terms by #3#, we get #x^2+2x-8=0#. This equation can be solved by factoring, because it is equal to #(x-2) \cdot (x+4)=0#.

The first example explains how to find the factorisation of a quadratic equation in standard form. The other two examples show how factorisation can be used to quickly solve a quadratic equation.

Write the expression #x^2+17\cdot x+72# as a product of linear factors.

#x^2+17\cdot x+72=# \((x+8)\cdot(x+9)\)

We are looking for numbers #p# and #q# such that the quadratic polynomial #x^2+17\cdot x+72# can be written as #(x-p)\cdot(x-q)#. If the absolute value of #p# is greater than #q#, we interchange them, so #|p|\le |q|#. We expand the brackets and compare the result with the original expression:
\[ x^2-(p+q)\cdot x+p\cdot q = x^2+17\cdot x+72\tiny\]
A comparison with #x^2+17\cdot x+72# gives \[
\lineqs{p+q &=& -17\cr p\cdot q &=& 72}\] If #p# and #q# are integers, they are divisors of #72#. We go through all possible divisors #p# with #p^2\le |72|# (which must be satisfied in view of #|p|\le |q|#) and in each case we calculate the sum of #p# and #q=\frac{72}{p}#:
\[\begin{array}{|r|c|l|}
\hline
p&q&{p+q}\\
\hline
1&72&73\\ \hline -1&-72&-73\\ \hline 2&36&38\\ \hline -2&-36&-38\\ \hline 3&24&27\\ \hline -3&-24&-27\\ \hline 4&18&22\\ \hline -4&-18&-22\\ \hline 6&12&18\\ \hline -6&-12&-18\\ \hline 8&9&17\\ \hline -8&-9&-17 \\
\hline
\end{array}\]
The line of the table with #p=-8# and #q=-9# is the only one with sum #-17#, hence, this is the answer:
\[x^2+17\cdot x+72=(x+8)\cdot(x+9)\tiny.\]

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