+0  
 
+2
84
10
avatar+75 

1. What is the largest value of k such that the equation 6x-x^2=k has at least one real solution?

 

2.Find the smallest real value of x such that x^2+6x+9=24.

 

3.Find the roots of x^2+12x+36+25=0
Note: The roots are not necessarily real.

 

4.Find all real solutions to x^2+4=100x^2+20x+1.

If you find more than one, then list the values separated by commas.

 Jan 26, 2019
 #1
avatar+98152 
+2

1. What is the largest value of k such that the equation 6x-x^2=k has at least one real solution?

 

Rearrange as    x^2 - 6x + k = 0

 

If this has at least one real solution, the discriminant must be ≥ 0

 

So

 

(-6)^2 -   4(1)(k) ≥ 0

 

36 - 4k ≥ 0

 

36 ≥ 4k 

 

9 ≥ k

 

So .....the max value  of k that produces at least one real solution is when k = 9

 

 

cool cool cool

 Jan 26, 2019
 #2
avatar+47 
+4

2. 

move all the numbers on one side so that the right side is left with 1

x^2+6x-15=0

use the quadratic formula to find smallest x can be

 

you should get something like

x = -3-2sqrt(6)

 

smiley

 Jan 26, 2019
 #4
avatar+98152 
0

Thanks, Jess !!!!

 

 

cool cool cool

CPhill  Jan 26, 2019
 #5
avatar+47 
0

no problem!

you are much better at explaining than me!

smiley

jess.shen2024  Jan 26, 2019
 #6
avatar+98152 
0

Don't know about that....LOL!!!!

 

 

cool cool cool

CPhill  Jan 26, 2019
 #7
avatar+75 
0

Thank you to both of you :D

Android4EVER  Jan 26, 2019
 #3
avatar+98152 
+4

2.Find the smallest real value of x such that x^2+6x+9=24.

 

Factor the left side

 

(x + 3)^2  = 24               take the negative root  [ since we want the smallest value ]

 

x + 3 = - √24       subtract 3 from both sides

 

x =  -√24 - 3  =    -2√6 - 3      and this is the smallest real value that makes the equation true

 

 

cool cool cool

 Jan 26, 2019
 #8
avatar+98152 
+2

4.Find all real solutions to x^2+4=100x^2+20x+1.

 

Rearrange as

 

99x^2 + 20x - 3   =  0

 

Using the quad formula.....we have

 

( -20 ±√ [ 20^2 - 4 * 99 * -3 ]  ) / (2 * 99)

 

( -20 ±√ 1588) / 198

 

(-20 ± 2√397 )/198 =

 

(-10 ±√397 ) / 99   .....and these are the two solutions

 

 

cool cool cool

 Jan 26, 2019
 #9
avatar+47 
+3

3.

Similarly to question 2, you should simplify all the like terms first

x^2+12x+61=0

 

in ax^2+bx=c you get a=1 b=12 c=61

 

use the quadratic formula to find the roots of the equation

 

solve this and you should get the 2 roots

 

x= -6+5i , x= -6-5i

 

smiley

 Jan 26, 2019
 #10
avatar+98152 
+1

Nice, Jess!!!

 

That should do it, Android!!!!!

 

 

cool cool cool

CPhill  Jan 26, 2019

20 Online Users

avatar
avatar
avatar
avatar