A circle of center (-3 , -2) passes through the points (0 , -6) and (a , 0). Find a.
The triangle bounded by the lines y = 0, y = 2x and y = -0.5x + k, with k positive, is equal to 80 square units. Find k.
When the polynomial P(x) = x3 + 3x2 -2Ax + 3, where A is a constant, is divided by x2 + 1 we get a remainder equal to -5x. Find A.
And would like to see work that can justify the answer. Thank you!
Hints:
(a)
find the radius r which corresponds to the distance between (-3,-2) and (0,-6). Then
(-3-a)^2+(-2-0)^2 = r^2
(b)
y=0 is horizontal (1)
y=2x (2) is a line through the origin sloping upwards
y=-0.5x+k (3) is a line sloping downwards with x-intercept: -0.5x+k=0, or x=2k.
You'll need to find the intercept between lines (2) and (3) and determine the height of the triangle.
(3)
This means that (x^2+1) divides P(x)-5x, or
P(x)-5x = (x^2+1)(x+3)=
=x^3+3*x^2+x+3
Therefore
P(x)=x^3+3*x^2+x+3 + 5x
=x^3+3*x^2+6x+3
=> A=-3
To find the value of a, we can use the properties of a circle.
1. First, we need to find the radius of the circle. The radius is the distance between the center and any point on the circle. In this case, we can calculate the distance between the center (-3, -2) and the point (0, -6) using the distance formula:
d = sqrt((x2 - x1)^2 + (y2 - y1)^2)
d = sqrt((0 - (-3))^2 + (-6 - (-2))^2)
d = sqrt(3^2 + (-4)^2)
d = sqrt(9 + 16)
d = sqrt(25)
d = 5
Therefore, the radius of the circle is 5 units.
2. Next, we can use the distance formula again to calculate the distance between the center (-3, -2) and the point (a, 0):
d = sqrt((x2 - x1)^2 + (y2 - y1)^2)
d = sqrt((a - (-3))^2 + (0 - (-2))^2)
d = sqrt((a + 3)^2 + 2^2)
d = sqrt(a^2 + 6a + 9 + 4)
d = sqrt(a^2 + 6a + 13)
Since this point is also on the circle, the distance must be equal to the radius, which is 5:
sqrt(a^2 + 6a + 13) = 5
a^2 + 6a + 13 = 25
a^2 + 6a - 12 = 0
Now we can solve this quadratic equation. We can use factoring, completing the square, or the quadratic formula to find the values of a. In this case, let's use the quadratic formula:
a = (-b ± sqrt(b^2 - 4ac)) / 2a
Plugging in the values:
a = (-6 ± sqrt(6^2 - 4(1)(-12))) / (2(1))
a = (-6 ± sqrt(36 + 48)) / 2
a = (-6 ± sqrt(84)) / 2
a = (-6 ± sqrt(4 * 21)) / 2
a = (-6 ± 2sqrt(21)) / 2
a = -3 ± sqrt(21)
Therefore, the values of a are -3 + sqrt(21) and -3 - sqrt(21).
To find the value of k in the triangle bounded by the lines y = 0, y = 2x, and y = -0.5x + k:
1. We need to find the intersection points between these lines.
y = 0 and y = 2x:
Setting y = 2x, we have 0 = 2x. Solving for x, we get x = 0.
y = 0 and y = -0.5x + k:
Setting y = -0.5x + k, we have 0 = -0.5x + k. Solving for x, we get x = 2k.
2. To find the area of the triangle, we can use the formula for the area of a triangle:
Area = (1/2) * base * height.
The base of the triangle is the distance between the intersection points x = 0 and x = 2k:
Base = 2k - 0 = 2k.
The height of the triangle is the y-coordinate of the intersection point x = k:
y = -0.5x + k
y = -0.5k + k
y = 0.5k.
Therefore, the area of the triangle is:
Area = (1/2) * (2k) * (0.5k)
= k^2.
Since the area of the triangle is given as 80 square units, we have:
k^2 = 80.
Taking the square root of both sides, we get:
k = ±sqrt(80).
We are given that k is positive, so we have:
k = sqrt(80).
Simplifying, we have:
k = 4sqrt(5).
Therefore, the value of k is 4sqrt(5).
To find the value of A in the polynomial P(x) = x^3 + 3x^2 - 2Ax + 3 when divided by x^2 + 1 and with a remainder of -5x:
1. When polynomial P(x) is divided by x^2 + 1, we can express it using synthetic division. Set up the division as follows:
_______________________
x^2 + 1 | x^3 + 3x^2 - 2Ax + 3
2. Synthetic division proceeds as follows:
- Divide the leading term of P(x) (x^3) by the leading term of the divisor (x^2): x^3 / x^2 = x.
- Multiply the divisor (x^2 + 1) by the divisor: (x)(x^2 + 1) = x^3 + x.
- Subtract this product from the first polynomial: (x^3 + 3x^2 - 2Ax + 3) - (x^3 + x) = 3x^2 - 2Ax + 3 - x = 3x^2 - 2Ax + (3 - x).
Thus, the intermediate polynomial after the first division is 3x^2 - 2Ax + (3 - x).
3. Next, we can continue dividing the intermediate polynomial by the divisor x^2 + 1.
- Divide the leading term of 3x^2 (in the intermediate polynomial) by the leading term of the divisor (x^2): 3x^2 / x^2 = 3.
- Multiply the divisor (x^2 + 1) by the divisor: (3)(x^2 + 1) = 3x^2 + 3.
- Subtract this product from the intermediate polynomial: (3x^2 - 2Ax + (3 - x)) - (3x^2 + 3) = -2Ax + (3 - x - 3) = -2Ax + (-x).
Thus, the intermediate polynomial after the second division is -2Ax + (-x).
4. Since we have a remainder of -5x, we can equate the remainder obtained (-2Ax + (-x)) with the remainder given (-5x):
-2Ax + (-x) = -5x.
5. Now, we can compare the like terms:
-2Ax - x = -5x.
6. Equating the coefficients of the x terms:
-2A - 1 = -5.
7. Solve for A:
-2A = -5 + 1.
A = (-5 + 1) / (-2).
A = -4 / -2.
A = 2.
Thus, the value of A is 2.
To summarize:
- The value of a can be found as a = -3 + sqrt(21) and a = -3 - sqrt(21).
- The value of k is 4sqrt(5).
- The value of A is 2.