Consider the equation
4x^2 – 16x + 25 = 0.
(a) Show how to compute the discriminant, b^2 – 4ac, and then state whether there is one real-number solution, two different real-number solutions, or two different imaginary-number solutions.
(b) Use the quadratic formula to find the exact solutions of the equation. Show work. Simplify the final results as much as possible.
(a) The equation is in the form ax^2 + bx + c
Use those values for b^2 - 4ac.
(-16)^2 - 4*4*25
= 256 - 400
= -144
The determinant is negative, so there will be two imaginary solutions.
(b) The quadratic equation is x = (-b +/- sqrt(determinant))/(2a)
Using the values of a, b, and c from the equation, we get
x = (16 +/- sqrt(-144))/(2*4)
x = (16 +/- 12i)/8
x = 2 +/- 3i/2
Would this be an appropriate answer for b?
x= 4+3i/2, 4-3i/2
How did you get 4 +/- 3i/2?
I figured 4*4 = 16. This is not right is it?
You divide by 8 at the end. 16/8 = 2
(a) To compute the discriminant, we can consider the equation in the form of ax^2 + bx + c = 0, where a = 4, b = -16, and c = 25.
The discriminant, denoted as Δ, is given by the formula: Δ = b^2 - 4ac.
Substituting the values, we have:
Δ = (-16)^2 - 4 * 4 * 25
= 256 - 400
= -144
Since the discriminant, Δ, is negative (-144), there are two different imaginary-number solutions to the equation.
(b) Now, let's use the quadratic formula to find the exact solutions of the equation:
The quadratic formula is given by: x = (-b ± √Δ) / (2a)
Substituting the values, we have:
x = (-(-16) ± √(-144)) / (2 * 4)
= (16 ± √144i) / 8
To simplify further, we can divide the numerator and denominator by 8:
= (2 ± √18i)
Therefore, the exact solutions of the equation 4x^2 – 16x + 25 = 0 are:
x = 2 + √18i
x = 2 - √18i