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How do I solve x^2+6x+8=0?

Ok, so we would call this a quadratic equation because it is written in the form of ax^2+bx+c=0 (in our case, a=1, b=6, c=8).

Luckily, this type of quadratic equation can be factorised, so we can solve it easily!

Ok, so we're trying to factorise x2+6x+8=0 so that it is in the form of (x+p)(x+q)=0.

Let's try expanding (x+p)(x+q)=0.

If you multiply the brackets together, you're left with:

x2+px+qx+pq=0

We can tidy this up a little bit to give us:

x2+(p+q)x+pq=0

This looks very similar to x2+6x+8=0, doesn't it?

Yes, it does! If we compare these two equations, we find out two things:

p+q=6

pq=8

So we're looking for two numbers which when they are multiplied by each other will give 8, and when they are added together will give 6. 

What we're left with is that p=4 and q=2 (or the other way round, it doesn't really matter).

Let's plug this back into our original equation: (x+p)(x+q)=0

Of course, now we have:

(x+4)(x+2)=0

This is much easier to solve than what we started with!

So in this case, either the first bracket is equal to 0 or the second bracket is equal to zero - this gives us two solutions for x.

Either x+4=0, meaning that x=-4

Or x+2=0, meaning that x=-2

So now we have the answers x=-2 or x=-4

Of course a quicker way to do this would be to look at x2+6x+8=0 and to find two numbers that are factors of 8 (they multiply together to make 8) and also add together to make 6.

William S. 11 Plus Maths tutor, A Level Maths tutor, 13 plus  Maths t...

12 months ago

Answered by William, a GCSE Maths tutor with MyTutor


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William S. 11 Plus Maths tutor, A Level Maths tutor, 13 plus  Maths t...
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