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.
\(\)
+129895 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
+129895 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 !!!!
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 \(\dfrac{\sqrt{x}}{x\sqrt{3}+\sqrt{2}} = \dfrac{1}{2x\sqrt{6}+4}\)
+129895 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
+32 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.
