Y^2 - 20y + 100 = 64
Looking for exact solutions. I know they are integers, but...is there a decimal solution?
Thanks
Marry me?
I doubt if you really want to marry a retired engineering professor who graduated from high school in 1955.
Yeah, it would be impossible to believe I'm near your age. I graduated in 1960..we never stop learning sir..
Thank you...
You are welcome :)
To solve the quadratic equation y^2 - 20y + 100 = 64, we can start by rearranging the equation:
y^2 - 20y + 100 - 64 = 0
This simplifies to:
y^2 - 20y + 36 = 0
To find the solutions, we can use the quadratic formula, which states that for a quadratic equation of the form ax^2 + bx + c = 0, the solutions can be found using:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = 1, b = -20, and c = 36. Plugging these values into the quadratic formula, we have:
y = (-(-20) ± √((-20)^2 - 4(1)(36))) / (2(1))
Simplifying further:
y = (20 ± √(400 - 144)) / 2
y = (20 ± √256) / 2
y = (20 ± 16) / 2
Now we have two possibilities:
y = (20 + 16) / 2 = 36 / 2 = 18
y = (20 - 16) / 2 = 4 / 2 = 2
Therefore, the two solutions to the equation y^2 - 20y + 100 = 64 are y = 18 and y = 2. Both of these values are integers.
To check if there is a decimal solution, we can substitute these values back into the original equation:
For y = 18:
18^2 - 20(18) + 100 = 64
324 - 360 + 100 = 64
64 = 64 (The equation is true)
For y = 2:
2^2 - 20(2) + 100 = 64
4 - 40 + 100 = 64
64 = 64 (The equation is true)
Since both solutions satisfy the equation, there are no decimal solutions to the equation y^2 - 20y + 100 = 64.
y^2 -20 y + 36 = 0
y = [ 20 +/- sqrt(400-144) ]/2
y = [20 +/- sqrt(256) ]/2
y = [ 20 +/- 16 ]/2
= 10 +/- 8
18 or 2
or
(y-2)(y-18) = 0