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  • 5 years ago

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.

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  1. anonymous
    • 5 years ago
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    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

  2. anonymous
    • 5 years ago
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    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.

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