# if x^2 + 9x + 20 = 0, what are the possible values of x?

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So x2 + 9x + 20 = 0

My preffered way of solving this equation is to factorise the equation. (Though I understand that different students may find other ways easier)

Factorisation is where the above equation is (x+a)(x+b) = 0

So if we times out (x+a)(x+b) we get

x2 + ax + bx + ab = 0
therefore
x2 + (a+b)x + ab = 0

Therefore we can equate this to the original question, so
x2 + 9x + 20 = x2 + (a+b)x + ab

so now we can see that
9 = a + b and
20 = ab

I would reccomend using trial and error (although I understand that different students may prefer other techniques).

So by trying for multiple values of a and b, we can see that they must equal 5 and 4.

Therefore
x2 + 9x + 20 = (x+5)(x+4) = 0

We know that the only way of producing a 0 through multiplication is through multiplying one number by another. Therefore we know that

x+5= 0 or x+4=0

Through rearranging these equations we can conclude that x must equal -4 or -5.

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