If the two roots of the equation AX & # 178; + BX + C = 0 (a ≠ 0) are X1 and X2, then X1 + x2 = - (B / a), x1x2 = C / A, prove

If the two roots of the equation AX & # 178; + BX + C = 0 (a ≠ 0) are X1 and X2, then X1 + x2 = - (B / a), x1x2 = C / A, prove

ax²+bx+c=0(a≠0)
From the root formula, we know that
x1=[-b+√(b²-4ac)]/(2a)
x2=[-b-√(b²-4ac)]/(2a)
So X1 + x2 = [- B + √ (B & # 178; - 4ac)] / (2a) + [- B - √ (B & # 178; - 4ac)] / (2a) = - B / A
x1x2=[-b+√(b²-4ac)]/(2a)*[-b-√(b²-4ac)]/(2a)=[(-b)²-(b²-4ac)]/(2a)²=4ac/(4a²)=c/a