Give exact and approximate solutions to three decimal places for y.
y^2-6y+9=64
If anyone can help with this problem I would really appriciate it. Thanks
Rewrite it in standard form
y^2 -6y -55 = 0
and then factor:
(y+11)(y-5) = 0
The answers are exact integers, so you don't have to worry about decimal places, unless that want you to write 5 as 5.000, etc.
Thank you for your help. So just to make sure its 11, 5 right. 5 is positive not negative??
To find the value of y, we can start by rearranging the equation:
y^2 - 6y + 9 = 64
Now, let's subtract 64 from both sides of the equation:
y^2 - 6y + 9 - 64 = 0
Simplifying further:
y^2 - 6y - 55 = 0
Now, we need to solve this quadratic equation. There are a few ways to do this, but one common method is to use the quadratic formula:
y = (-b ± √(b^2 - 4ac)) / (2a)
Here, a = 1, b = -6, and c = -55. Substituting these values into the quadratic formula:
y = (-(-6) ± √((-6)^2 - 4(1)(-55))) / (2(1))
Simplifying:
y = (6 ± √(36 + 220)) / 2
y = (6 ± √256) / 2
y = (6 ± 16) / 2
Now, let's consider both possible solutions:
For y = (6 + 16) / 2 = 22 / 2 = 11
For y = (6 - 16) / 2 = -10 / 2 = -5
So, the two solutions for y are 11 and -5.
These are exact solutions. If you are asked to find approximate solutions to three decimal places, you can use a calculator to evaluate the square root if required and round the answer.