SOLUTION: give exact and approximate solutions to three decimal places for y^2-12y+36=100

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Question 211375: give exact and approximate solutions to three decimal places for
y^2-12y+36=100
Found 2 solutions by jim_thompson5910, rapaljer:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

y%5E2-12y%2B36=100 Start with the given equation.


y%5E2-12y%2B36-100=0 Get every term to the left side.


y%5E2-12y-64=0 Combine like terms.


Notice that the quadratic y%5E2-12y-64 is in the form of Ay%5E2%2BBy%2BC where A=1, B=-12, and C=-64


Let's use the quadratic formula to solve for "y":


y+=+%28-B+%2B-+sqrt%28+B%5E2-4AC+%29%29%2F%282A%29 Start with the quadratic formula


y+=+%28-%28-12%29+%2B-+sqrt%28+%28-12%29%5E2-4%281%29%28-64%29+%29%29%2F%282%281%29%29 Plug in A=1, B=-12, and C=-64


y+=+%2812+%2B-+sqrt%28+%28-12%29%5E2-4%281%29%28-64%29+%29%29%2F%282%281%29%29 Negate -12 to get 12.


y+=+%2812+%2B-+sqrt%28+144-4%281%29%28-64%29+%29%29%2F%282%281%29%29 Square -12 to get 144.


y+=+%2812+%2B-+sqrt%28+144--256+%29%29%2F%282%281%29%29 Multiply 4%281%29%28-64%29 to get -256


y+=+%2812+%2B-+sqrt%28+144%2B256+%29%29%2F%282%281%29%29 Rewrite sqrt%28144--256%29 as sqrt%28144%2B256%29


y+=+%2812+%2B-+sqrt%28+400+%29%29%2F%282%281%29%29 Add 144 to 256 to get 400


y+=+%2812+%2B-+sqrt%28+400+%29%29%2F%282%29 Multiply 2 and 1 to get 2.


y+=+%2812+%2B-+20%29%2F%282%29 Take the square root of 400 to get 20.


y+=+%2812+%2B+20%29%2F%282%29 or y+=+%2812+-+20%29%2F%282%29 Break up the expression.


y+=+%2832%29%2F%282%29 or y+=++%28-8%29%2F%282%29 Combine like terms.


y+=+16 or y+=+-4 Simplify.


So the solutions are y+=+16 or y+=+-4


Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
This happens to be a Quadratic Equation that FACTORS! There are TWO ways to solve it. I'm not sure which way you want to solve it.

First, you can factor the left side of the equation and take the square root of each side:
y^2-12y + 36=100
(y-6)^2 = 100
y-6 = +10 or - 10

y-6=10
y=16

y-6=-10
y-6+6=-10+6=-4

The other way is to set the equation equal to zero, like this:
y^2-12y+36-100=0
y^2 -12y -64 =0

This just happens to factor also (but it doesn't ALWAYS work this way!)
(y-16)(y+4) = 0
y =16 or y =-4

R^2

Dr. Robert J. Rapalje
Seminole Community College
Altamonte Springs Campus