SOLUTION: Find all solutions of the equation |x^2 - 30x - 1| = 30 + |6x - 2| + |x^2 - 20x - 4|.

Algebra ->  Absolute-value -> SOLUTION: Find all solutions of the equation |x^2 - 30x - 1| = 30 + |6x - 2| + |x^2 - 20x - 4|.      Log On


   

Question 1209411: Find all solutions of the equation |x^2 - 30x - 1| = 30 + |6x - 2| + |x^2 - 20x - 4|.
Found 3 solutions by Edwin McCravy, KMST, ikleyn:
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!

abs%28x%5E2+-+30x+-+1%29+=+30+%2B+abs%286x+-+2%29+%2B+abs%28x%5E2+-+20x+-+4%29

Get 0 on the right:

abs%28x%5E2+-+30x+-+1%29+-+30+-+abs%286x+-+2%29+-+abs%28x%5E2+-+20x+-+4%29=0

Graph the left side in your TI-84 calculator. 

The "abs" feature is found this way: MATH NUM 1:abs(

Put it in Y1 like this:

abs(x^2 - 30x - 1) - 30 - abs(6x - 2) - abs(x^2 - 20x - 4)

Set the window Xmin=-10, Xmax=25, Ymin=-40, Ymax=60

Your graph will look something close to this:



Use the zero feature of the calculator, and you'll find three solutions

The "zero" feature is found 2ND TRACE 2:zero

x = -7.25,   x = 7.75,  and x = 21.464225

Edwin

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
Before graphing calculators, it would have been a more complicated

The right hand is greater than 30 except for all values of x
30%2Babs%286x-2%29%2Babs%286x-2%29%3E=30 for all of real values because
abs%286x-2%29%3E=0 for all real numbers,
with abs%286x-2%29=6x-2%3E=0 for x%3E=1%2F3, and abs%286x-2%29=-6x%2B2 otherwise.
abs%28x%5E2+-+20x+-+4%29%3E=0 for
x%3C=10-sqrt%28104%29=approximately-0.198039027 and x%3E=10%2Bsqrt%28104%29=approximately20.198039027.
x%5E2+-+20x+-+4%3C0 and abs%28x%5E2+-+20x+-+4%29%3E0 for
for x such that 10-sqrt%28104%29%3Cx%3C10%2Bsqrt%28104%29

For the left hand side, x%5E2+-+30x+-+1+=0, has as solutions x=15+%2B-+sqrt%28226%29
For values of x%3C=15+-+sqrt%28226%29=approximately-0.033296378 and x%3E=15+%2B+sqrt%28226%29=approximately30.033296378,
x%5E2+-+30x+-+1%3E=0 and abs%28x%5E2+-+30x+-+1%29=x%5E2+-+30x+-+1.
For 15+-+sqrt%28226%29%3Cx%3C15+%2B+sqrt%28226%29 abs%28x%5E2+-+30x+-+1%29=-x%5E2+%2B+30x+%2B+1
Those limiting values are approximately -0.033296378 and 30.033296378.

For x%3C=10-sqrt%28104%29=approximately-0.198039027, 6x-2%3C0, while x%5E2+-+30x+-+1+%3E0 and x%5E2+-+20x+-+4%3E=0
and the equation turns into
x%5E2+-+30x+-+1+=+-6x%2B2+%2B+x%5E2+-+20x+-+4
-30x+%2B+6x+%2B+20x+=+30+%2B+2+-+4+%2B+1
-4x+=+29 --> x=29%2F%28-4%29=highlight%28-7.25%29

For -1%3Cx%3C1 x%5E2+-+30x+-+1 is decreasing from 30 at x=-1 to a minimum 0f -1 at x=0 and then increasing to 30 at x=1 and abs%28x%5E2+-+30x+-+1%29 is less than the value of 30%2Babs%286x-2%29%2Babs%28x%5E2+-+20x+-+4%29%3E=30
=
For 1%3C=x%3C=10%2Bsqrt%28104%29, abs%286x-2%29=6x-2%3E0, x%5E2-20x-4%3C=0, and x%5E2-30x-1%3C0,
so abs%28x%5E2-20x-4%29=-x%5E2%2B20x%2B4%29 and abs%28x%5E2-30x-1%29=-x%5E2%2B30x%2B1
The equation turns into
-x%5E2%2B30x%2B1=30%2B6x-2%2B-x%5E2%2B20x%2B4
30x-6x-20x=30-2%2B4-1
4x=31 --> x=31%2F4=highlight%287.75%29

For 10%2Bsqrt%28104%29%3Cx%3C15%2Bsqrt%28226%29 abs%286x-2%29=6x-2%3E0 while
abs%28x%5E2-20x-4%29=x%5E2-20x-4%3E0 and x%5E2-30x-1%3C0, so abs%28x%5E2-30x-1%29=-x%5E2%2B30x%2B1
turning the equation into -x%5E2%2B30x%2B1=30%2B6x-2%2Bx%5E2-20x-4
0=x%5E2-30x-1%2B30%2B6x-2%2Bx%5E2-20x-4
0=2x%5E2-44x%2B23 --> x=%2844+%2B-+sqrt%281752%29%29%2F4=11+%2B-+sqrt%28438%29%2F2
x=11-sqrt%28438%29%2F2=approximately0.536 does not comply with 10%2Bsqrt%28104%29%3Cx%3C15%2Bsqrt%28226%29
x=highlight%2811%2Bsqrt%28438%29%2F2%29=approximately21.464225 is a solution between 10%2Bsqrt%28104%29=approximately20.198039027 and 15%2Bsqrt%28226%29=approximately30.033296378

For x%3E=15%2Bsqrt%28226%29 abs%286x-2%29=6x-2%3E0, abs%28x%5E2-20x-4%29=x%5E2-20x-4%3E0 , and abs%28x%5E2-30x-1%29=x%5E2-30x-1%3E=0 ,
turning the equation into x%5E2-30x-1=30%2B6x-2%2Bx%5E2-20x-4
-30x-6x%2B20x=30-2-4%2B1
-16x=25 --> x=25%2F%28-16%29=-1.5625 is not a solution complying with x%3E=15%2Bsqrt%28226%29=approximately30.033296378

Answer by ikleyn(52858) About Me  (Show Source):
You can put this solution on YOUR website!
.

For problems like this one, there is a standard methodology/strategy for solving.

You should subdivide the number line by sections (= intervals).
The division point are the points where the participating functions under absolute values
(parts of the equation) change their signs.

Then for each interval, from minus infinity to infinity, you write an equation
using each function with its native sign as it should be, dictated by the absolute
value rule at this interval.

Then you solve the updated equation and check if the solution does belong to this particular
interval under the consideration.

If the particular solution does belong to this particular interval, then the found solution
is the solution to the original equation.
If it does not belong, then the found solution is not a solution to the original equation.

As you complete going from the left -infinity to the right infinity, you complete the solution.

The rest is just a technique and the arithmetic.


I think, if a student does understand it and demonstrates his/her understanding, it should be just
enough for the teacher to accept the assignment.


As the problem is twisted in this post, it is twisted TOO much and teaches nothing.


A teacher who assigns such tasks, should be checked for adequacy.