How do you do quadratic equations?

Answer 1

You can solve a quadratic equation in lots of different ways depending on what type it is.

A quadratic equation is any equation in the form ax2 + bx + c = 0 (rather than y = mx + c (linear equation), ax3 + bx2 + cx + d = 0 (cubic equation) or ax4 + bx3 + cx2 + dx + e = 0, etc.). An example of a quadratic equation is 3x2 + 5x - 4 = 0. a = 3, b = 5 and c = -4. You can also get quadratic equations in forms such as i) x2 + 4x + 2 = 0, ii) x2 - 4x = 0, iii) x2 + 2 or iv) 5x2 + x + 1 = 2x - 4. They are still quadratic because i) a = 1, ii) c = 0, iii) b = 0 and iv) b = 1 and if you do some rearranging, you get x2 - x + 5 = 0. In general, in quadratic equation, the x-term's highest power is 2.

The easiest type of quadratic equation to solve is x2 + yx. For example, if you have x2 + 3x = 0, you solve it by doing this:

x(x + 3) = 0 Factorise

If x multiplied by x + 3 = 0 then either x = 0 or x + 3 = 0

Therefore, x = 0 or -3

In almost all quadratic equation, you end up with two answers because when you square a number, that number and the same negative number (e.g. 5 and -5) both give the same answer, therefore you can have 2 answers. On a quadratic graph, it is a curve going up then down or visa versa, therefore, it must cross the x-axis twice (unless b2 - 4ac is a negative number because then you have to start using the square roots of negative numbers and therefore i), this is the same answer as solving it mathematically and is how you solve equations graphically. The only time that you have one answer is if after you factorise, you end up with something like (x - 2)(x - 2) = 0 where both brackets are the same. In this case the answer is therefore only 2.

The next type of quadratic equation is the full equation. Here is a worked example:

x2 + 3x + 10 = 4 - 2x

x2 + 5x + 6 = 0 Rearrange (add 2x and take 4 from both sides)

You now have to factorise the equation. You do this by finding two numbers that add to give b (the number in bold (+ 5) and multiply to give the number in c (the number in italics (+ 6). In this case, the answer is + 2 and + 3 as 2 + 3 is 5 and 2 x 3 is 6. If it were x2 - 7x + 12, the answer would be - 3 and - 4 as - 3 - 4 = - 7 and - 3 x - 4 = 12.

(x + 2)(x + 3) = 0 Factorise

x + 2 = 0 or x + 3 = 0 so x = - 2 or - 3

The next type of equation is a complete equation when a is not 1. Here is an example:

6p2 - 12p - 48 = 0

For this question, you have to think a bit harder as the p-term will be greater than 1 when in the brackets so you have to think about what numbers will multiply to make a (in this case 1 x 6 and 2 x 3 = 6) and then apply it so that you can find an answer. Lets guess that you need 2p and 3p so you need to find two numbers that multiply to make c (- 48) and add when multiplied by 2 or 3 to make - 12. Lets also guess that you need - 6 and 8, - 6 x 8 = - 48 (good so far) but - 6 x 2 + 8 x 3 = 12 and it's not -12 when you change which numbers you multiply by. So lets make another guess that you need - 12 and 4. - 12 x 4 = -48 (again, good so far) and -12 x 3 + 4 x 2 = - 12 so we therefore find that:

(3p - 12)(2p + 4) = 0 Factorise

So 3p - 12 = 0 or 2p + 4 = 0 so 3p = 12 or 2p = - 4 so p = 4 or - 2

There is no mathematical process to working out the factorising part, you just have to use trial and error.

There is however another way of solving all the quadratic equations mentioned. You can use the quadratic equation formula:

x = (- b + sprt (b2 - 4ac)) / 2a or (- b - sqrt (b2 - 4ac)) / 2a

If you are not used to shorthand, sqrt means square root. Notice the subtle difference between the two, adding or subtracting the sqrt.

Here is an example:

2p2 - 6p + 4 = 0

So a = 2, b = - 6 and c = 4

p = (- (- 6) + sqrt ((- 6)2 - 4 x 2 x 4)) / 2 x 2 or (- (- 6) - sqrt ((- 6)2 - 4 x 2 x 4)) / 2 x 2

p = (6 + sqrt (36 - 32)) / 4 or (6 - sqrt (36 - 32)) / 4

p = (6 + sqrt (4)) / 4 or (6 - sqrt (4)) / 4

p = (6 + 2) / 4 or (6 - 2) / 4

p = 8 / 4 or 4 / 4 so p = 2 or 1

This may look harder but is much quicker, easier and guaranteed to give you an answer but the other methord is longer, harder and may not always give an answer (e.g. if the answer is 78 / 13 or - 178 / 101).

The final type that I will cover is simultaneous equations. One is linear and the other quadratic. The two answers are the points on a graph where the lines will cross.

(1) y = x + 2

(2) y = x2 + 6x + 8

Knowing this infomation, substitute x + 2 into y = x2 + 6x + 8.

x + 2 = x2 + 6x + 8 Substitute

x2 + 5x + 6 = 0 Rearrange

(x + 2)(x + 3) Factorise

x = - 2 or - 3 Work out x

Now that you know x, it's just a simple matter of shoving it into the linear equation and doing a nice little bit of simple maths!

y = - 2 + 2 or - 3 + 2

y = 0 or - 1

So x = - 2, y = 0 or when x = - 3, y = - 1

And that is where the maths will end. This is all of quadratics. I haven't shown you quadratic graphs but they're quite simple, you just have to plot every point you can so if told to plot a quadratic graph where x is between - 3 and 3, plot - 3, - 2, - 1, 0, 1, 2 and 3. The point where it crosses the x-axis is the answer! I hope this has been helpful to you. Have fun doing some maths!