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Algebra Difficulty 5.8 AIME, harder Prove it Ukraine

It is known, that x1,x2,x3x_1, x_2, x_3 are distinct real numbers.

a) x2,x3x_2, x_3 are zeros of the function f1(x)=x2+p1x+q1f_1(x) = x^2 + p_1x + q_1; x3,x1x_3, x_1 are zeros of the function f2(x)=x2+p2x+q2f_2(x) = x^2 + p_2x + q_2; x1,x2x_1, x_2 are zeros of the function f3(x)=x2+p3x+q3f_3(x) = x^2 + p_3x + q_3. Does the function f(x)=f1(x)+f2(x)+f3(x)f(x) = f_1(x) + f_2(x) + f_3(x) always have zeros?

b) x2,x3x_2, x_3 are zeros of the function f1(x)=a1x2+b1x+c1f_1(x) = a_1x^2 + b_1x + c_1; x3,x1x_3, x_1 are zeros of the function f2(x)=a2x2+b2x+c2f_2(x) = a_2x^2 + b_2x + c_2; x1,x2x_1, x_2 are zeros of the function f3(x)=a3x2+b3x+c3f_3(x) = a_3x^2 + b_3x + c_3. Does the function f(x)=f1(x)+f2(x)+f3(x)f(x) = f_1(x) + f_2(x) + f_3(x) always have zeros?

Solution

a) We write our function in the form
f(x)=(x−x2)(x−x3)+(x−x1)(x−x3)+(x−x1)(x−x2), f(x) = (x - x_2)(x - x_3) + (x - x_1)(x - x_3) + (x - x_1)(x - x_2),
WLOG, x1<x2<x3x_1 < x_2 < x_3. Then f(x2)=(x2−x1)(x2−x3)<0f(x_2) = (x_2 - x_1)(x_2 - x_3) < 0, which is equivalent to the existence of the roots of the function.

b) Consider the following three functions:
f1(x)=x2+x,f2(x)=x2−x,f3(x)=1−x2, f_1(x) = x^2 + x, \quad f_2(x) = x^2 - x, \quad f_3(x) = 1 - x^2,
They have the roots 0,−10, -1, 0,10, 1 and −1,1-1, 1 respectively. Their sum
f(x)=x2+x+x2−x+1−x2=x2+1, f(x) = x^2 + x + x^2 - x + 1 - x^2 = x^2 + 1,
does not have zeros.

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