Question

Solve \[m^{2} - 14m + 13 = 0\] by first comparing it with the standard form \[am^{2} + bm + c = 0\]. Which set of coefficients is obtained?

Options

• \[a = 1, \; b = 14, \; c = 13\]

• \[a = -14, \; b = 1, \; c = 13\]

• \[a = 1, \; b = -14, \; c = 13\]

• \[a = 1, \; b = -14, \; c = -13\]

Answer 1

Comparing \[m^{2} - 14m + 13 = 0\] with \[am^{2} + bm + c = 0\] gives \[a = 1\], \[b = -14\], and \[c = 13\]. The sign of the middle term is carried directly into \[b\], so \[b = -14\], not \[+14\].