02.01 Solving Quadratic Equations by Factorising

VIDEO LESSON

INTERACTIVE SELF-STUDY

A quadratic equation can be written in the form \(ax^2+bx+c=0\), where \(a, b\) and \(c\) are real constants, and \(a≠0\).
i. What kind of equation would we have if \(a=0\)? Show Answer

Quadratic equations can have two, one, or no real solutions

Some (but not all) quadratic expressions can be factorised as a product of two linear factors, that is, can be expressed in the form \(ax^2+bx+c≡(px+q)(rx+s)\)
(Sometimes we are very fortunate and \(p,q,r\) and \(s\) will be integers)

• Hence we can sometimes rewrite the quadratic equation \(ax^2+bx+c=0\) as \((px+q)(rx+s)=0\)

By factorising the quadratic in this way we can make some deductions

First, note that if \(m×n=0\), then either \(m=0\) or \(n=0\) (or both).
In other words, if the product of two terms is \(0\), then at least one of the terms is equal to \(0\)

We have \((px+q)×(rx+s)=0\)
i. Hence what two deductions can we make? Show Answer

• If \((px+q)(rx+s)=0\), then the solutions for \(x\) are \(\displaystyle x=-\frac{q}{p}\) or \(\displaystyle x=-\frac{s}{r} \)
• A solution to an equation is sometimes called the root of the equation

If the two linear factors are identical, that is, if the quadratic equation is factorised as \((px+q)(px+q)=(px+q)^2\) then the quadratic has exactly one solution that appears twice.

• This is called a repeated root or a double root

To solve a quadratic equation remember to rearrange it to the form \(ax^2+bx+c=0\) before factorising

Worked Example 1 – (GRADE E)
Solve the equation \(x^2−x−6=0\) by factorising Show Answer

Example 2 – (GRADE D)
a. Solve the equation \(6x^2+x−15=0\) Show Answer

Example 3 (Walkthrough) – (GRADE D)
Consider the equation \(x^2=7x\)

We may be tempted to divide both sides by \(x\), however in doing so we lose a solution.

• You should never divide an equation by a variable/term because that variable/term could be equal to \(0\), and division by \(0\) is not allowed!
  (If you are told or know for certain that the term cannot be equal to \(0\), then division by that term is allowed)
• Instead you must bring all terms to one side by addition/subtraction and then factorise the equation

Hence solve the equation \(x^2=7x\). Show Answer

Notice that we have a quadratic equation which resulted in two solutions. Had we divided both sides by \(x\) we would have obtained a linear equation, \(x=7\), with only one solution. So we can see that premature division has made us lose a solution.

ALTERNATIVE FORMS OF QUADRATIC EQUATIONS

Consider the equation: \((px+q)^2=r\).
For an equation in this form, it would actually be longer to solve by expanding, rearranging and factorising.
It is in fact more straightforward to simply square root both sides of the equation:
\((px+q)^2=r \iff px+q=±\sqrt{r}\)

• Remember when square rooting an equation you must consider both the positive and negative square roots

Example 4 – (GRADE D)
Solve the equations
a. \((2x+1)^2=9 \) Show Answer

END OF LESSON!

You can now answer:
EDEXCEL Pure Year 1 Ex 2A – Q1 to Q4