ES3: Quadratic equations
The general form of a quadratic equation is A X squared plus B X plus C equals zero where A is not equal to one.
This tip sheet looks at solutions to quadratic equations using the “null factor law”.
General form
A equation can be rearranged to the form:
\(ax^{2}+bx+c=0\) where
\(a\neq0\).
Examples
(using \(ax^{2}+bx+c=0\) determine \(a\) , \(b\) and \(c\))
- \(5x^{2}-3x+9=0\) \(\quad\quad a=5,\,\) \(\ b=-3,\,\) \(\ c=9\)
- \(x^{2}=5x-4\ \ \ \Rightarrow\) \(\ \ \ x^{2}-5x+4=0\quad\quad\) \(a=1,\,\) \(\ b=-5,\,\) \(\ c=4\)
- \(x=\frac{3}{2x}\ \ \ \) \(\Rightarrow\ \ \ \) \(2x^{2}=3\) \(\ \ \ \Rightarrow\ \ \ \) \(2x^{2}-3=0\qquad\) \(a=1,\,\) \(\ b=0,\,\) \(\ c=-3\)
Factorisation
If the equation can be factorised then the ‘null factor law’ can be used to find the solutions:
Null factor law uses the simple idea:
If \(m\times n=0\), then \(m=0\) and / or \(n=0\)
If the product of two or more factors is zero then any one of the individual factors may be zero and provide a solution for the equation.
Example 1 (find all possible values of y)
\[\begin{alignat*}{1} y^{2} & =5y\\ y^{2}-5y & =0\,(\mathrm{\mathrm{rearrange\,}to\,form})\\ y(y-5) & =0\,(\mathrm{rearrange})\\ \mathrm{for\ this\ to\ be\ true}\\ y=0\,\ \mathrm{or}\,\, & (y-5)\ \,\mathrm{must\ equal\ zero,\ \ \ (null\,factor\,law})\\ \therefore\mathbf{\,y=0\,\,\mathrm{or}\,\,} & \mathbf{y=5}\\ \\ \mathrm{check\ by\ substututing\ back\ to\ } & y^{2}=5y\\ \mathrm{If}\,\, & y=0,\ \ 0^{2}=5\times0\mathrm{\,(checked)}\\ \mathrm{If}\,\, & y=5,\ \ 5^{2}=5\times5\,\mathrm{(checked)}\\ y^{2} & =5y\\ \mathrm{Left\,Hand\ Side}\ & \mathrm{=Right\,Hand\ Side}\\ \end{alignat*}\]
Example 2 (find all possible values of x)
\[\begin{alignat*}{1} x^{2}-5x+4 & =0\\ (x-4)(x-1) & =0\,\mathrm{\ (factorise)}\\ (x-4)=0\,\ \,\mathrm{or}\,\ \, & (x-1)=0\,\ \ \mathrm{(null\,factor\,law})\\ \mathbf{\therefore\ x=4\,\,\mathrm{or}}\,\, & \mathbf{x=1}\\ \\ \mathrm{check\ by\ substututing\ back\ to\ } & x^{2}-5x+4=0\\ \mathrm{If}\,\, & x=4,\ \ 4^{2}-5\times4+4=0\mathrm{\,(checked)}\\ \mathrm{If}\,\, & x=1,\ \ 1^{2}-5\times1+4=0\mathrm{\,(checked)}\\ \end{alignat*}\]
Example 3 (find all possible values of p)
\[\begin{alignat*}{1} p^{2}+10p+25 & =0\\ (p+5)(p+5) & =0\,\mathrm{\ \ (factorise)}\\ (p+5) & =0\ \ \,\mathrm{(null\,factor\,law})\\ \therefore\mathbf{\ p} & \mathbf{=-5}\\ \\ \mathrm{check\ by\ substututing\ back\ to\ } & p^{2}+10p+25=0\\ \mathrm{If}\ p=-5\,\,\mathrm{check} & (-5)^{2}+10\times(-5)+25=0\\ \end{alignat*}\]
Example 4 (find all possible values of m)
\[\begin{alignat*}{1} 4m^{2}-49 & =0\\ (2m+7)(2m-7) & =0\,\mathrm{(factorise\,by\,difference\,of\,squares)}\\ (2m+7)=0\,\mathrm{or}\,\, & (2m-7)=0\,\mathrm{(null\,factor\,law)}\\ \mathbf{\therefore\ m=\frac{-7}{2}\ \ \,\mathrm{or}}\,\ \, & \mathbf{m\mathbf{=\frac{7}{2}}}\\ \\ \mathrm{check\ by\ substututing\ back\ to\ } & 4m^{2}-49=0\\ \end{alignat*}\]
Example 5 (find all possible values of x)
\[\begin{alignat*}{1} x=\frac{-6}{1-2x}\,\,\mathrm{provided}\,\, & x\neq1/2\\ x-2x^{2} & =-6\mathrm{\ \ \ (multiplying\ both\ sides\ by}\ 1-2x)\\ 2x^{2}-x-6 & =0\,\ \ \mathrm{(rearrange\,to\,form\,\,}ax^{2}+bx+c=0)\\ (2x+3)(x-2) & =0\,\ \ \mathrm{(factorise)}\\ \mathbf{\therefore\ x=\frac{-3}{2}\,\mathrm{\ \ or\ }\,\,x} & \mathbf{=2\,}\ \ \,\mathrm{(solving\,using\,null\,factor\,law)}\\ \end{alignat*}\]
Exercises
Solve the following quadratic equations:
- \(\ \ x^{2}-6x+8=0\)
- \(\ \ x^{2}+2x-3=0\)
- \(\ \ 2x^{2}-3x-2=0\)
- \(\ \ 6-z-z^{2}=0\)
- \(\ \ 2x^{2}+7x=15\)
- \(\ \ 11p=3(2p^{2}+1)\)
\[\begin{array}{ccc} 1. & \ \ x=4, & \ \ x=2\\ 2. & \ \ x=-3, & \ \ x=1\\ 3. & \ \ x=-1/2, & \ \ x=2\\ 4. & \ \ z=-3, & \ \ z=2\\ 5. & \ \ x=3/2, & x=-5\\ 6. & \ \ p=1/3, & \ \ p=3/2 \end{array}\]
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