4C - Solving quadratic equations
Quadratic equations can be solved by:
- Factorising and using the null factor law.
- Using a CAS calculator.
- Using the general quadratic formula.
■ The null factor law state that if \(a\times b=0\), then \(a=0\), \(b=0\) or \(a=b=0\).
The null factor law can be used to solve simple quadratic equations. To solve quadratic equations:
- Manipulate the equation so that the quadratic equation is equal to zero; that is, \(ax^2+bx+c=0\).
- Factorisise the quadratic expression.
- Use the null factor law to determine the solution(s).
4C - Example 1: Solving quadratic equations using the null factor law
Solve the following equation for \(x\) \[(x-5)(2x-3)=0\]
|
4C - Example 1: Video solution
4C - Example 1: Practice
Question 1: Solve the equation \((x-2)(x+6)=0\) for \(x\). Question 2: Solve the equation \((2x+1)(x-4)=0\) for \(x\). Question 3: Solve the equation \((3x-2)(x-9)=0\) for \(x\). Question 4: Solve the equation \((x+p)(2x+q)=0\) for \(x\). 4C - Example 1: Solutions
Question 1: \[x=2, x=-6\] Question 2: \[x=-\frac{1}{2}, x=4\] Question 3: \[x=\frac{2}{3},x=9\] Question 4: \[x=-p, x=-\frac{q}{2}\] |
4C - Example 2: Solving quadratic equations using the null factor law
Solve the following equation for \(x\) \[x^2-6x+8=0\]
|
4C - Example 2: Video solution
4C - Example 2: Practice
Question 1: Solve the equation \(x^2+7x+10=0\) for \(x\). Question 2: Solve the equation \(x^2-x-6=0\) for \(x\). Question 3: Solve the equation \(-x^2+12x-32=0\) for \(x\). 4C - Example 2: Solutions
Question 1: \[x=-2, x=-5\] Question 2: \[x=-2, x=3\] Question 3: \[x=4, x=8\] |
4C - Example 3: Solving quadratic equations using the null factor law
Solve the following equation for \(x\) \[2x^2+9x=-4\]
|
4C - Example 3: Video solution
4C - Example 3: Practice
Question 1: Solve the equation \(3x^2+11x-4=0\) for \(x\). Question 2: Solve the equation \(2x^2+5x-7=0\) for \(x\). Question 3: Solve the equation \(-4x^2=7x-15\) for \(x\). 4C - Example 3: Solutions
Question 1: \[x=-4, x=\frac{1}{3}\] Question 2: \[x=-\frac{7}{2}, x=1\] Question 3: \[x=-3,x=\frac{5}{4}\] |
4C - Example 4: Solving quadratic equations (CAS)
Solve each of the following equations for \(x\) using CAS:
(a) \(15x^2+11x-14=0\) (b) \(x^2-18x+26=0\) |
4C - Example 4: Video solution
4C - Example 4: Practice
Question 1: Solve the equation \(21x^2-29x-10=0\) for \(x\) using a CAS calculator. Question 2: Solve the equation \(x^2-6x+4=0\) for \(x\) using a CAS calculator. 4C - Example 4: Solutions
Question 1: \[x=-\frac{2}{7}, x=\frac{5}{3}\] Question 2: \[x=3-\sqrt{5}, x=3+\sqrt{5}\] |
-
Additional Exercises
-
Topic Worksheets
-
Other Resources
<
>
Question 1
Solve each of the following equations for \(x\): (a) \(x^2-12=0\) (b) \(x^2+16x+64=0\) (c) \(-2x^2+3x=-27\) Question 2 [NZL, 2012] Mark solves the equation \[\frac{x^2-5x+6}{x^2+x-6}=4\] His working is shown below, labelled line (1) to line (5). (1) \(x^2-5x+6=4x^2+4x-24\) (2) \(3x^2+9x-30=0\) (3) \(3(x^2+3x-10)=0\) (4) \(3(x+6)(x-2)=0\) (5) \(x=-6, x=2\) Is Mark's answer correct? Fully justify your answer. |
Solutions:
Question 1 (a) \(x=2\sqrt{3},x=-2\sqrt{3}\) (b) \(x=-8\) (c) \(x=-3, x=\frac{9}{2}\) Question 2 No, Mark's answer is incorrect. He has made a mistake factorising in line (4). This should be \(3(x+5)(x-2)=0\) giving \(x=-5\) and \(x=2\) as possible solutions. However, \(x=2\) cannot be a solution as the denominator would equal \(0\). Therefore, the only solution is \(x=-5\). |
Solving quadratic equations - Worksheet A
|