Question Detail

A quadratic equation has only two roots. The mention of three roots suggests a misprint. Possibly the roots are 1 and -6. Then sum = -b/a = 1 + (-6) = -5, product = c/a = 1(-6) = -6. Taking a=1, b=5, c=-6. Then a+b+c = 1+5-6=0. But option A is 0. However, if the roots are 1, 1, and -6, it's not a quadratic. If the intended was 1 and -6, a+b+c=0. If the intended was 1, -6, and something else, not possible. Given the options, -7 is also there. If roots are 1 and 1, then sum=2, product=1. Then a=1, b=-2, c=1, a+b+c=0. So 0 appears. If roots are 1 and -6, a+b+c=0. If roots are 1, 1, -6, it's cubic. The most plausible answer from the options is 0. However, the image shows "1, 1 and -6" which is three roots, so it must be a cubic. For a cubic ax³+bx²+cx+d=0 with roots 1,1,-6, sum of roots = -b/a = 1+1-6=-4 => b/a=4. Sum of pairwise products = c/a = (11)+(1-6)+(1-6)=1-6-6=-11. Product = -d/a = 11-6=-6 => d/a=6. Let a=1, b=4, c=-11, d=6. Then a+b+c = 1+4-11 = -6. Not an option. So likely the intended was a quadratic with two roots. Given the options, 0 is the most consistent.