Question 137369

In this question we know that the number have to be consecutive. So we call them: x, x+1, x+2 (so if we find x to be, say, 7, then our numbers are 7, 8, 9, which are consecutive). Now we need to find the sum of their squares, which is given to be 77. Now we have {{{x^2 + (x+1)^2 + (x+2)^2 = 77}}}. Now we need to simplify, and solve for x. Remember that {{{(x+c)^2=x^2 +2*c*x+c^2}}}, which is the simplified, or nice, version of foil for expanding a square. So we get {{{x^2 + (x^2 +2*x + 1) + (x^2 + 4*x + 4) = 3*x^2 +6*x + 5 = 77}}}. Now we set our equation up so we can use the formula {{{x=(-b +- sqrt(b^2 - 4*a*c))/(2*a)}}}, where are equation must be in the form {{{a*x^2+b*x+c = 0}}}. So we have {{{3*x^2 + 6*x -72 = 0}}} or {{{x^2 + 2*x - 24 = 0}}} after dividing both sides by 3. Now if you want, you can plug in a=1, b=2, and c= -24 into our formula above, or you can factor by saying what are the factors of -24, that add up to +2. Well we have 6*4=24, and 6 + -4=2, so we get {{{(x-4)*(x+6)=0}}}, which you can double check the FOIL. Now this equation can only equal zero if one of the two factors on the left hand side are equal to zero. In other words, we must have either x-4=0, or x+6=0. Now we see that x=4 or -6. In any case we get our three consecutive integers as 4, 5, 6, or -6, -5, -4, which you can see that adding the squares of them will give you the same answer for both sets of number (since one set is just the opposite, or negative of the other).