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

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) (Show Source): You can put this solution on YOUR website!

Start with the given equation.

Get every term to the left side.

Combine like terms.

Notice that the quadratic is in the form of where , , and

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

Start with the quadratic formula

Plug in , , and

Negate to get .

Square to get .

Multiply to get

Rewrite as

Add to to get

Multiply and to get .

Take the square root of to get .

or Break up the expression.

or Combine like terms.

or Simplify.

So the solutions are or

Answer by rapaljer(4671) (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
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