Use the quadratic formula to find the exact solutions of the equation. Show work. Simplify the final results as much as possible.
4x^2 – 16x + 25 = 0.
The exact question was asked by Pamela yesterday. I have copied down my solution. For more information, view the entire thread.
(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
To find the exact solutions of the equation 4x^2 – 16x + 25 = 0 using the quadratic formula, we first need to identify the coefficients of the quadratic equation. In this case, the coefficients are:
a = 4
b = -16
c = 25
Now, we can substitute these values into the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the coefficients, we have:
x = (-(-16) ± √((-16)^2 - 4(4)(25))) / (2(4))
Simplifying further:
x = (16 ± √(256 - 400)) / 8
x = (16 ± √(-144)) / 8
We can see that the expression inside the square root (√(-144)) yields a negative number, which means that the equation has no real solutions. This is because the discriminant (b^2 - 4ac) is negative.
So, there are no exact real solutions to the equation 4x^2 – 16x + 25 = 0.
In complex numbers, we can rewrite the equation as:
x = (16 ± √(144) i) / 8
Further simplifying:
x = (16 ± 12i) / 8
Finally,
x = 2 ± 3i
Thus, the exact solutions of the equation 4x^2 – 16x + 25 = 0 are x = 2 + 3i and x = 2 - 3i.