Find three consecutive integers whose sum of their squares is 194. Write an equation, and solve the equation.Your equation MUST be an algebra equation with a variable.

Question

anonymous · Answer

x^2+(x+1)^2+(x+2)^2 = 194  ===> 3x^2+(x^2+3x+1)+(x^2+4x+4) =194 ===> combining ===> 3x^2+6x+5=194 ===> x^2+2x-63 =0 ===> (x-7)(x+9)=0 ===> x=7 or x=-9 ===> x=7 ==> the numbers are 7 , 8 and  9

anonymous · Answer

If the three integers are consecutive, then that means they are of the form$$x, x+1, x+2$$or more simply$$x-1,x,x+1$$If the sum of their squares is 194, then you can write:$$(x-1)^2+x^2+(x+1)^2=194$$From there, expand the brackets, and you'll end up with quite a simple equation which can be solved to find x (as well as x - 1, and x + 1).

Take note of the fact that the question asks for "three consecutive $\underline{integers}$", which means they can be positive or negative. In fact, solving for x will get you two sets of numbers, which are equally valid.