Solve the equation x^2-x-56=0

The first step in this solution is to factorise the equation we have been given into the form (ax+b)(cx+d)=0, where a, b, c and d are constants which we need to determine. If we were to expand the brackets we get teh equation acx^2+dax+bcx+bd=0. By gathering like terms (which means terms corresponding to similar powers of x) we get the equation acx^2+(da+bc)x+bd=0. Comparing this rearranged form to the original equation we see that ac=1, da+bc=-1 and bd=-56. We can then solve these simultaneous equatins to find the solution. If ac=1 then a=1 and c=1 (there are other solutions but it is better to assumer we are only dealing with integers in the first instance). da+bc=-1 and so by substituting a and c we get d+b=-1. We can also see that bd=-56. Therefore we have to find two numbers that add to -1 and multiply to -56. Since these are both negative numbers we can see that either b or d is negative. First consider what two numbers multiply to 56: 7 and 8. Thus either -7 and 8 or -8 and 7 are possible solutions. -7+8=1 which is not what we need, but -8+7=-1 wich works for our equation. Therefore a=1, b=-8, c=1 and d =7 (NB if a and c differ in value then b and d must be determined with reference to our origianl factorising equation. In the case where a=1 and c=1 the values of b and d can be switched with no effect). Substituting this into our equation we get (x-8)(x+7)=0. Therefore, either x-8=0 or x+7=0 and hence the solutions to this equation are x=8 and x=-7. You should then substitue your answers into the question to confirm that they are correct.