Give exact and approximate solutions to three decimal places.
(x-4)^2=20
(x-4)(x-4)=20
x^2-8x+16=20
x^2-8x-4=0
Now use the quadratic formula.
To find the solutions to the given equation, we can start by expanding the equation:
(x-4)^2 = 20
Expanding the equation gives us:
(x-4)(x-4) = 20
x^2 - 8x + 16 = 20
Now, let's rearrange the equation to the standard quadratic form:
x^2 - 8x + 16 - 20 = 0
x^2 - 8x - 4 = 0
To find the solutions, we can apply the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this equation, a = 1, b = -8, and c = -4.
Substituting these values into the quadratic formula:
x = (-(-8) ± √((-8)^2 - 4(1)(-4))) / (2(1))
x = (8 ± √(64 + 16)) / 2
x = (8 ± √80) / 2
x = (8 ± 8.944) / 2
Now, let's calculate the values of x:
x1 = (8 + 8.944) / 2 ≈ 8.472
x2 = (8 - 8.944) / 2 ≈ -0.472
So, the solutions to the equation (x-4)^2 = 20, rounded to three decimal places, are approximately x ≈ 8.472 and x ≈ -0.472.