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# Hey New Topic i need help with?

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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

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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

Jan 26, 2019
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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)

Jan 26, 2019
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Thanks, Jess !!!!

CPhill  Jan 26, 2019
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no problem!

you are much better at explaining than me!

jess.shen2024  Jan 26, 2019
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CPhill  Jan 26, 2019
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Thank you to both of you :D

Android4EVER  Jan 26, 2019
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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

Jan 26, 2019
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4.Find all real solutions to x^2+4=100x^2+20x+1.

Rearrange as

99x^2 + 20x - 3   =  0

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

( -20 ±√ 1588) / 198

(-20 ± 2√397 )/198 =

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

Jan 26, 2019
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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

Jan 26, 2019
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Nice, Jess!!!

That should do it, Android!!!!!

CPhill  Jan 26, 2019