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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 \(2ix^2+x+3i=0.\)

 Oct 15, 2019
 #1
avatar+106536 
+1

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)

 

 

 

cool cool cool

 Oct 15, 2019
edited by CPhill  Oct 15, 2019
 #3
avatar+190 
+1

Oh i didn't think about the discriminant, thanks!

Oofrence  Oct 15, 2019
 #2
avatar+106536 
+1

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)

 

 

cool cool cool

 Oct 15, 2019
 #4
avatar+106536 
+1

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

 

 

 

cool cool cool

 Oct 15, 2019

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