SOLUTION: y^2-12y+36=49 c^2+16c+64=15

Question 112305: y^2-12y+36=49
c^2+16c+64=15
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Start with the given equation

Subtract 49 from both sides

Factor the left side (note: if you need help with factoring, check out this solver)

Now set each factor equal to zero:

or

or Now solve for y in each case

So our solutions are or

Notice if we graph (just replace y with x) we get

and we can see that the graph has roots at and , so this verifies our answer.

#2

Start with the given equation

Subtract 15 from both sides

Let's use the quadratic formula to solve for c:

Starting with the general quadratic

the general solution using the quadratic equation is:

So lets solve ( notice , , and )

Plug in a=1, b=16, and c=49

Square 16 to get 256

Multiply to get

Combine like terms in the radicand (everything under the square root)

Simplify the square root (note: If you need help with simplifying the square root, check out this solver)

Multiply 2 and 1 to get 2

So now the expression breaks down into two parts

or

Now break up the fraction

or

Simplify

or

So these expressions approximate to

or

So our solutions are:
or

Notice when we graph (just replace c with x), we get:

when we use the root finder feature on a calculator, we find that and .So this verifies our answer

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