Solving Quadratics by Factorising

Quadratic equations have an x^2 term and no higher power of x, such as x^2-x-12=0. To solve one means to find the values of x that make it true.

Most quadratic equations have two solutions. Here they are x=4 and x=-3. Try x=4: 4^2-4-12=16-4-12=0, so it works.

In the exam you are usually asked to solve an equation such as x^2+3x-10=0, for 3 marks, near the end of a paper. You can get a mark just for the correct brackets, so always write them down. If a question says hence, use your answer to the part before.

If two numbers multiply to give 0, at least one of them must be 0. For example, if 7\times n=0, then n must be 0

Brackets work in the same way. If (x-4)(x+6)=0, then either the first bracket is 0 or the second bracket is 0. So either x-4=0 or x+6=0

Solve each one: x-4=0 gives x=4, and x+6=0 gives x=-6. So the solutions are x=4 or x=-6

Notice that each solution has the opposite sign to the number in its bracket: (x-4) gives 4, and (x+6) gives -6. To check, put each solution into its bracket. It should make 0

Most questions give the quadratic multiplied out, in the form x^2+bx+c=0. You factorise it into two brackets first, then solve each bracket.

• Check that one side of the equation is 0
• Find two numbers that multiply to give c and add to give b.
• Write the equation as two brackets equal to 0
• Set each bracket equal to 0 and solve it.

If the two numbers are p and q, then

x^2+bx+c=(x+p)(x+q)

where p\times q=c and p+q=b. The solutions are x=-p and x=-q

Sometimes both brackets are the same. x^2-8x+16=(x-4)(x-4), so x^2-8x+16=0 has only one solution, x=4

In x^2+5x=0 there is no number on its own. Both terms contain x, so take x out as a common factor: x(x+5)=0

Now the two things multiplied are x and x+5. So x=0 or x+5=0, which gives x=0 or x=-5

Don't divide both sides by x. That leaves x+5=0, and you lose the solution x=0

In x^2-25=0 there is no x term, and 25 is a square number. This is a difference of two squares, which factorises as (x+5)(x-5)

So x+5=0 or x-5=0, which gives x=-5 or x=5. Both work, because 5^2=25 and (-5)^2=25 too.

In general, for a positive number k,

x^2-k^2=(x+k)(x-k)

so the solutions of x^2-k^2=0 are x=k and x=-k. Don't forget the negative one.

The solutions of x^2+bx+c=0 are where the graph of y=x^2+bx+c crosses the x-axis, because y=0 all along the x-axis.

Solving gives the same values: x^2-x-6=(x+2)(x-3)=0, so x=-2 or x=3. If a question asks you to solve, give the values of x, not coordinates.

In a problem, x often stands for a length, an age or a number of people. These can't be negative, so if one solution is negative, reject it and use the other. Then answer the question that was asked.

• Stopping once you have factorised. (x-8)(x+5)=0 is not the answer; set each bracket equal to 0 and write down the values of x.
• Getting the signs of the solutions wrong. (x-8)(x+5)=0 gives x=8 or x=-5, not x=-8 or x=5; check that each value makes its bracket 0.
• Giving only one solution. Trying numbers might find x=8 but miss x=-5; factorise to find both.
• Solving it like a linear equation. Adding or subtracting terms on both sides can't get x on its own when there is an x^2 term; factorise instead.
• Dividing both sides by x. In x^2-12x=0 this loses x=0; take out x as a common factor instead.
• Forgetting the negative solution. x^2-49=0 has two solutions, x=7 and x=-7