if x^2 + 9x + 20 = 0, what are the possible values of x?
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 getx2 + ax + bx + ab = 0 therefore x2 + (a+b)x + ab = 0Therefore we can equate this to the original question, so x2 + 9x + 20 = x2 + (a+b)x + abso now we can see that 9 = a + b and 20 = abI 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 thatx+5= 0 or x+4=0 Through rearranging these equations we can conclude that x must equal -4 or -5.
