Question

find the zeros of the quartic polynomial 18x²-2x-11​

Answer 1

Answer:

f(x) = 18x² - 2x - 11

Let, α and β are two roots of this polynomial.

D= b² - 4ac = 4 - 4(18)(-11) = 4 + 792 = 796

√D = 28 (approximately)

Now,

α = (b² + √D) /2a and, β = (b² - √D) /2a

α = 4 + 28 / 32 = 32/32 = 1

β = 4 - 28 / 32 = -24/32 = -3/4