SOLUTION: Consider the equation 4x^2 – 4x + 5 = 0.
(i) Compute the discriminant, b2 – 4ac, and then state whether one real-number solution, two different real-number solutions, or two di
Algebra ->
Radicals
-> SOLUTION: Consider the equation 4x^2 – 4x + 5 = 0.
(i) Compute the discriminant, b2 – 4ac, and then state whether one real-number solution, two different real-number solutions, or two di
Log On
Question 147444: Consider the equation 4x^2 – 4x + 5 = 0.
(i) Compute the discriminant, b2 – 4ac, and then state whether one real-number solution, two different real-number solutions, or two different imaginary-number solutions exist.
(ii) Use the quadratic formula to find the exact solutions of the equation.
You can put this solution on YOUR website! 4x^2 – 4x + 5 = 0
4x^2 + (-4)x + 5 = 0
ax^2 + bx + c = 0
a = 4
b = -4
c = 5
(i) Compute the discriminant, b2 – 4ac, and then state whether one real-number solution, two different real-number solutions, or two different imaginary-number solutions exist.
disc. > 0 ~ two real
disc. = 0 ~ one real
disc. < 0 ~ two imaginary
(ii) Use the quadratic formula to find the exact solutions of the equation.
(