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How many real solutions are there for x in the following equation: (x - 5x + 12)^2 + 1 = -|x|

 

Evaluate \(\log_3 27\sqrt3\). Express your answer as an improper fraction.

\(\)

 Aug 4, 2023
 #1
avatar+128732 
+1

Here's the second one

 

log3 27sqrt (3)  = 

 

log3 27 + log3 sqrt (3) = 

 

log 3 3^3  + log 3 3^(1/2) =

 

(3) log 3 3  + (1/2) log3 3  =

 

(3) (1) + (1/2) (1)  =

 

3 + 1/2  =

 

7/ 2

 

cool cool cool

 Aug 4, 2023
 #2
avatar+128732 
+1

Here's a logical (graphical) solution for the first one

 

Looking at the left side of the equation :

 

(x -5x + 12)^2 + 1  =

 

(12 - 4x)^2 + 1  =

 

144 - 96x + 16x^2 + 1 =

 

16x^2 - 96x + 145      (1)

 

This is a parabola that turns upward    with the x coordinate of the vertex at   - (-96)/ [ 2 * 16]  =

 

96 / 32  =  3

 

Plugging this value  back into (1) resulrs in yhe y coordinate of the  vertex = 

 

16 (3)^2  - 96(3)   + 145  =  1

 

So....the LOWEST point on this graph =  ( 3,1)

 

Meanwhile....the graph of  -abs (x)  reaches its HIGHEST point at (0,0)

 

Thus the graph of the left side of this equation never intersects the graph of the right side of the equation....therefore.....there are no real  solutions

 

CORRECTED ANSWER   !!!!

 

 

cool cool cool

 Aug 4, 2023
edited by CPhill  Aug 4, 2023
edited by CPhill  Aug 4, 2023
 #3
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Ummm I have another few questions:

1. What is the sum of all integer solutions to \(|n| < |n-3| < 9 \) ?

 

2. Find x, given that \(\dfrac{\sqrt{x}}{x\sqrt{3}+\sqrt{2}} = \dfrac{1}{2x\sqrt{6}+4}\)

 Aug 4, 2023
 #5
avatar+128732 
+1

abs (n) < abs ( n -3) < 9       we can write

 

sqrt (n^2) < sqrt [ ( n -3)^2 ] < 9

 

Taking the first part

 

sqrt (n^2) < sqrt [ (n -3)^2 ]    square both sides

 

n^2 < (n -3)^2

 

n^2 < n^2 -6n + 9

 

0 < -6n + 9

 

6n < 9

 

n < 9/6

 

n < 3/2   so..... n takes on all integer values from  -infinity to 1   → (1)

 

Taking the second part

 

sqrt [ (n -3)^2 ]  < 9      square both sides

 

(n -3)^2  < 81

 

n^2 - 6n + 9 < 81

 

n^2 - 6n - 72 < 0

 

(n - 12) ( n + 6) < 0

 

This will be true when   n is on the imterval of (-6 ,12).....so n takes on all integer values from -5 to 11  →   (2)

 

However the intersection of (1) and (2) are the integers    -5, -4, -3 , -2, -1 , 0 , 1

 

The sum of these =   -14 

 

 

cool cool cool

CPhill  Aug 4, 2023
 #4
avatar+32 
0

For your second post:

1. n must be a negative number, as all positive values of n would be greater than n-3. n must also be greater than -6, as -6-3=-9, and the absolute value of that would be nine, so 0>n>-6, so the possible integer values of n are 0, -1, -2, -3, -4, and -5. Add that up, and you get -15.

 Aug 4, 2023
 #6
avatar+32 
+1

oh yeah I forgot the 1. Thanks, Cphill!

jonathanldong  Aug 4, 2023
 #7
avatar+128732 
0

No prob....your answer was pretty close  !!!

 

 

cool cool cool

CPhill  Aug 4, 2023

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