College Algebra Help. Quadratics and coordinates

Question

2 Questions from a college algebra homework assignment, that I am stuck on. Please help me understand these. 

 

1. Find the numbers k for which the quadratic equation x^2 +kx+k=0 has no real solutions.

 

2. In the same coordinate system, sketch the graphs of the following two circles: x^2-4x+y^2+2y-4=0 and x^2-4x+y^2+2y-11=0 and find the area of the region bounded by the two circles (leave it exact). 

 

Thank you for your help. 

Answer 1

For the first part, use the quadratic equation and analyze the term in the sqrt (b^2-4ac). In this example, it looks like a=1 and b=c=k, so the term to analyze is: k*k-4k. When this term is less than zero, then the roots will not be real, so set k*k-4k<0 and solve for k. It looks like the result is when k< 4, but you should verify. Hope this helps. The summary is that you are looking for what values of k will cause the term in sqrt of the quadratic equation to be <0.

Answer 2

I'm still struggling with Number 1.  :D

Number 2 isn't so bad, so let's begin with it.

The first equation is x^2-4x+y^2+2y-4=0.

We add 4 to both sides, and have (and I am spacing the equation out for a reason):

x^2 - 4x      + y^2 + 2y      = 4

Now, I take half of the coefficient of -4, (-2), and square it:  (-2)^2 = 4.  Similarly, half of the coefficient of y, 2, is 1, and 1^2 is 1.  So, I add 4 and 1 to both sides:

x^2 - 4x + 4 + y^2 + 2y + 1 = 4 + 4 + 1

 

The reason I had spaced these out is to add these numbers in where they correspond, next to the x's or to the y's.

Now, we have two perfect squares on the left, so we factor them and have:

(x-2)^2 + (y+1)^2 = 9.

 

Recall that the equation for a circle is (x-h)^2 + (y-k)^2 = r^2, so this is a circle centered at (2, -1), with a radius of 3.

Similarly, the other equation results in a circle centered at (2,-1), but with a radius of 4.

Now, we have to find the area of the region bounded by these two concentric circles.  I am assuming that means the area inside the bigger one, but outside the smaller one.

Since a=(pi)r^2, to find the area outside the smaller circle but inside the bigger circle is the area of the bigger circle minus the area of the smaller circle:

A-a = (pi)R^2 - (pi)r^2 = (pi)(R^2-r^2) = (pi)(16-9) = (pi)*7 = 7(pi).

:)

Answer 3

First, if you post each question separately, you'll get more/better help on each one.

 

 

 

Now, look at the first question, and think about graphing the answer.  if you have an x^2 term, and a kx term, and a k term, under what circumstances will the answer be exactly zero, and cross the x axis?

 

x^2 will always be non-negative - and it's only zero at x=0.   

 

Under what circumstances is it possible that

 

x^2 = -kx - k

 

So, what do you know about k, if the equation has a zero?

 

 

 

 

Second problem - the general form for a circle is (x - xcenter)^2 + (y - ycenter)^2 = radius^2

 

the simple way to find the x and y coordinates of the center is to cut in half the coefficient of the x and y terms (-4x and +2y)

 

(x - 2)^2 + (y+1)^2 = r^2  

 

You can multiply that all out if you want, I'm going to shortcut.  The x will throw off +4, and the y will throw off +1, but we ended up with -4 on the left side of the equal sign, so we have a number of r^2 that is 4+4+1 = 9.  Therefore,  in this case, r turns out to be 3.  You have a circle, centered at (2,-1) with radius 3.

 

Similarly, the other circle turns out to be centered at (2,-1) with a radius of 4.   

 

"The region bounded by the two circles" is subject to interpretation, but I'd bet he's talking about the doughnut that's within the 4-radius and not within the 3-radius.  Plug pi-r-squared for radius 4, subtract pi-r-squared for radius 3, adn you have an answer.