What is.. .................squared?

Question

x = 2y +15

im suppossed to plug this in into x^2 + y^2 = 50

and get a quadratic

so ( 2y + 15 ) ^ 2 +y ^2 = 50

what is the final quadratic equation ?

Answer 1

(2y + 15)^2 + y^2 = 50

(4y^2 + 60y + 225) + y^2 = 50

5y^2 + 60y + 225 - 50 = 0

5y^2 + 60y + 175 = 0

Answer 2

4y^2 + 60y + 225 +y^2= 50

5y^2 + 60y + 175 = 0

Answer 3

To find the final quadratic equation, you need to solve for both x && y from the above standard equation of a circle.  The standard equation of a circle is: x^2 + y^2 = r^2.  In this case, it is x^2 + y^2 = squarerootof(50)^2, which is thus equal to the same standardized expression.  Well, there is no need to solve for x because it is always defined for you.  Solving for y yields, y = x - 15 / 2. Thus, the final quadratic equation is:

x^2 + y^2 = 50 = (2y + 15)^2 + (x - 15 / 2)^2 = 50.  To simply the 2nd binomial factor, separate the numerators to obtain

(2y + 15)^2 + (x/2 - 15/2)^2 = 50.

Also, to simply the above equation, it is easier to leave y^2 by itself.  That is, now you are left with

(2y + 15)^2 + y^2 = 50 = (2y + 15)(2y + 15) + y^2 = 50.  By leaving y unsolved, you have left:

5y^2 + 60y + 225 = 50.  Thus, the absolute final quadratic equation is: 5y^2 + 60y + 175 = 0.

J.C

Answer 4

5y^2 + 60y + 175 = 0

Answer 5

When squaring a binomial like 2y + 15, just imagine multiplying it by itself.  Then you can use FOIL (first, outer, inner, last)

Example:

(2y + 15)² + y² = 50

(2y + 15)(2y + 15) + y² = 50

(4y² + 30y + 30y + 225) + y² = 50

(4y² + 60y + 225) + y² = 50

Combine the y² terms:

5y² + 60y + 225 = 50

Subtract 50 from both sides:

5y² + 60y + 175 = 0

Note:  You could also divide both sides by 5 to simplify it:

y² + 12y + 35 = 0

Then factoring is easy:

(y + 5)(y + 7) = 0

y = -5 or y = -7

Plug that back into the first equation to solve for x:

x = 2(-5) + 15 = 5

or

x = 2(-7) + 15 = 1

Answer:

The intersection points of the line and the circle are:

(5, -5) and (1, -7)