+220 1. The graphs of \(y = x^2 + 2x + 7\) and \(y = 6x + b\) intersect at only one point. What is the value of x?
I think you should do x^2 + 2x + 7 = 6x + b, but i don't know how to continue
2. Find all pairs (x, y) such that \( x^2 + y^2 - 2x + 4y + 5 = 0.\)
I dont' know how to start this problem at all
3. Find the solutions of x in the equation \(2ix^2+x+3i=0.\)
+128448 I know a couple
y = x^2 + 2x + 7
y = 6x + b
So
x^2 + 2x + 7 = 6x + b subtract 6x + b from both sides
x^2 - 4x + (7 - b) = 0
If we have only one intersection point......the discriminant will = 0
So
(-4)^2 - 4(1)(7 - b) = 0 simplify
16 - 28 + 4b = 0
-12 + 4b = 0
4b = 12
b = 3
Look at the graph here : https://www.desmos.com/calculator/2eit022nxs
The intersection point is (2, 15)
+128448 2)
Rearrange as
x^2 -2x + y^2 + 4y = -5 complete the square on x, y
x^2 - 2x + 1 + y^2 + 4y + 4 = -5 + 1 + 4
(x - 1)^2 + (y + 2)^2 = 0
This is known as a "degenerate" circle ( circle with radius = 0)
It has a "center" of ( 1, - 2) = (x, y)
+128448 3)
(2i)x^2 + x + 3i = 0 factor out the 2i
2i ( x^2 + (1/2i)x + 3/2) = 0 divide both sides by 2i
x^2 + (1/2i)x + 3/2 = 0
x^2 + (1/(2i))x = - 3/2 complete the square on x
Take (1/2) of (1 /(2i)) = (1/(4i))
Square this = (1/(4i)) (1/ (4i)) = 1/(16i^2) = 1/ -16 = -1/16
Add this to both sides
x^2 + (1/(2i))x - 1/16 = -3/2 - 1/16 factor the left.....simplify the right
(x + (1/(4i))^2 = -25/16 take both roots
x + (1/(4i) ) = ±(5/4)i subtract (1/(4i)) from both sides
x = ±(5/4)i - (1/(4i)) multiply top/bottom of the last fraction by i
x = ±(5/4)i - i / (4i^2)
x = ±(5/4)i - (i / (-4) )
x = ±(5/4)i + i/4
x = ±(5/4)i + (1/4)i
x = ±(5/4)i + (1/4)i
x = (5/4)i - (1/4) i = (4/4)i = i
or
x = (-5/4)i - (1/4)i = (-6/4)i = (-3/2)i
So.....the solutions are
x = (-3/2)i and x = i
