Solve quadratic equations using the quadratic formula
Prerequisites from previous chapters:
Surds
VIDEO LESSON
INTERACTIVE SELF-STUDY
Because all quadratic expressions can be expressed as a completed square, we can find the general solution to quadratic equations in terms of their coefficients \(a,b\) and \(c\). This gives us the quadratic formula.
i. Complete the square for the expression \(ax^2+bx+c\), leaving your answer in the form \(a(x+p)^2+q\), where \(q\) is written as a single fraction
Here is the answer: \(\displaystyle =a\left[x^2+\frac{b}{a}x\right]+c\) \(\displaystyle = a\left[\left(x+\frac{b}{2a}\right)^2-\left(\frac{b}{2a}\right)^2\right]+c\) \(\displaystyle =a\left[\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a^2}\right]+c\) \(\displaystyle =a\left(x+\frac{b}{2a}\right)^2-a\cdot\frac{b^2}{4a^2}+c\) \(\displaystyle =a\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a}+c\) \(\displaystyle =a\left(x+\frac{b}{2a}\right)^2 + \left(\frac{4ac-b^2}{4a}\right)\)
So we obtain that if \(ax^2+bx+c=0\) , then \(\displaystyle x= \frac{-b\pm\sqrt{b^2-4ac}}{2a}\) This is called the quadratic formula. It will not be given in your formulae booklet
Since the quadratic formula is the end result of using completing the square to solve a quadratic equation, just go straight to the quadratic formula whenever you are unable to factorise a quadratic equation
Example 1 – (GRADE E) Solve \(16x^2+24x−89=0\) using the quadratic formula
Here is your hint: Identify \(a,b\) and \(c\) and substitute them into the formula
Here is the answer: \(a=16\), \(b=24\), \(c=-89\) \(\sqrt{b^2-4ac}=\sqrt{(24)^2-4(16)(-89)}=\sqrt{6272}=56\sqrt{2}\) \(\displaystyle x= \frac{-24\pm56\sqrt{2}}{32}= \frac{-3\pm7\sqrt{2}}{4}\)
END OF LESSON!
You can now answer: EDEXCEL Pure Year 1 Ex 2B – All questions