Question

Determine the type and number of solutions of 4x^2-5x+1=0.
A. Two real solutions
B. One real solution
C. Two imaginary solutions

Answer 1

4x² - 5x + 1 = 0
a = 4; b = - 5, c = 1
Δ = b² - 4.a.c
Δ = (-5)² - 4.4.1
Δ = 25 - 16
Δ = 9

x = - b ± √Δ / 2.a
$x = \frac{-5\± \sqrt{9} }{2*4}$
$x = \frac{-5\±3}{8}$
$x' = \frac{-5-3}{8} = \frac{-8}{8} = -1$
$x' = \frac{-5+3}{8} = \frac{-2}{8} (\div2) = \frac{-1}{4}$


A. Two real solutions (x' = -1 and x'' = -1/4)