Question

SOLVE IT IF U ARE POLYNOMIALS GENIUS..... CHALLENGE!​

Answer 1

We know that
if roots are equal than
${b}^{2} - 4ac = 0$
${(c - a)}^{2} - 4(b - c)(a - b) = 0 \\ {c}^{2} - 2ac + {a}^{2} - 4(ab - {b}^{2} - ac + bc) = 0 \\ {c}^{2} - 2ac + {a}^{2} - 4ab + 4 {b}^{2} + 4ac - 4bc = 0 \\ {c}^{2} + {a}^{2} - 4ab + 4{b}^{2} + 2ac - 4bc = 0 \\ {a}^{2} + ac - 2ab + 4 {b}^{2} - 2ab - 2bc + {c}^{2} + ac - 2bc = 0 \\ a(a + c - 2b) - 2b( - 2b + c + a) + c(c + a - 2b) = 0 \\ {(a + c - 2b)}^{2} = 0 \\ a + c - 2b = 0 \\ a + c = 2b \: \\ proved$