Hi again cleanncrazi!!
1)2m^2-m-15=0
This kind of problems are solved using the quadratic formula:
A polynomial a.x^2 + b.x + c , has two possible roots that can be
found using the Quadratic Formula:
x1 = [-b + sqrt(b^2 - 4.a.c)] / (2.a)
and
x2 = [-b - sqrt(b^2 - 4.a.c)] / (2.a)
For additional reference see the following pages:
"Quadratic Polynomials: The Quadratic Formula":
http://www.sosmath.com/algebra/factor/fac08/fac08.html
"Quadratic Equation Calculator":
http://www.1728.com/quadratc.htm
The two possible solutions or ROOTS are: 3 and -2.5
m1 = [1 + sqrt(1^2 - 4.2.(-15)] / (2.2) =
= [1 + sqrt(1 + 120)] / 4 =
= [1 + sqrt(121)] / 4 =
= [1 + 11] / 4 =
= 12/4 =
= 3
m2 = [1 - sqrt(1^2 - 4.2.(-15)] / (2.2) =
= [1 - sqrt(1 + 120)] / 4 =
= [1 - sqrt(121)] / 4 =
= [1 - 11] / 4 =
= -10/4 =
= -2.5
----------------------------------------------------------
2)2x^2+5x-1=0
This problem is similar to the previous one:
x1 = [-5 + sqrt(5^2 - 4.2.(-1))] / (2.2) =
= [-5 + sqrt(25 + 8)] / 4 =
= [-5 + sqrt(33)] / 4 =
= 0.18614
x2 = [-5 - sqrt(5^2 - 4.2.(-1))] / (2.2) =
= [-5 - sqrt(25 + 8)] / 4 =
= [-5 - sqrt(33)] / 4 =
= -2.6861
-----------------------------------------------------------
3)3x-x^2>0
3.x - x^2 > 0 ==> x.(3-x) > 0
Then both x and (3-x) must have the same sign:
Case x > 0:
If x > 0 then must be
3-x > 0 ==> 3 > x
Then must be 0 < x < 3
Case x < 0:
If x < 0 then must be
3-x < 0 ==> 3 < x
This situation is not possible (x<0 and 3<x at the same time)
Then the only solution is: 0 < x < 3
-----------------------------------------------------------
4)x^2+25<10x
x^2 + 25 < 10x <==> x^2 -10x + 25 < 0
Now finding the roots using the quadratic formula we will have that
the two roots are the same (test this with the quadratic calculator
page):
x1 = x2 = 5 ;
Then we have that:
x^2 -10x + 25 = (x-5)*(x-5) = (x-5)^2
NOTE: remember that if you have a polynomial ax^2 + bx + c with roots
x1 and x2, then:
ax^2 + bx + c = a.(x-x1).(x-x2)
The problem can be rewritten as:
(x-5)^2 < 0 ,
and since all real number different to zero raised to the second power
is possitive (i.e. greater than zero) this situation is not possible.
-----------------------------------------------------------
5)3/x+2>2/x-1
3/(x+2) > 2/(x-1)
It is obvious that x cannot be -2 or 1 .
Consider (x > 1):
Then (x-1)>0 and (x+2)>0 , so we can say that the original inequation
is similar to the following:
3(x-1) > 2(x+2) <==> 3x-3 > 2x+4 <==>
<==> 3x-2x > 4+3 <==>
<==> x > 7
Then must be x > 1 (our original condition) and x > 7 ; then for this
first case the solution is:
-----
x > 7
-----
Now consider x < 1, then (x-1) < 0;
If x < -2 then (x+2) < 0, and (x-1) < 0 ;
3/(x+2) > 2/(x-1) <==> 3(x-1)/(x+2) < 2 <==>
<==> 3(x-1) > 2(x+2)
The inequation changes the direction two times and we obtain the same
inequation of the first case, that has a solution x>7, but we start
considering x < 1 AND x < -2, so this case is not possible.
Now consider x < 1 (then (x-1)< 0 ) and x > -2 (then (x+2)> 0 ):
We have -2 < x < 1 and (x-1)< 0 and (x+2)> 0 ; then:
3/(x+2) > 2/(x-1) <==> 3(x-1)/(x+2) < 2 <==>
<==> 3(x-1) < 2(x+2) <==>
The inequation changes the direction once.
<==> 3x-3 < 2x+4 <==>
<==> 3x-2x < 4+3 <==>
<==> x < 7
For this case we have x < 7 , and the original condition -2 < x < 1;
both conditions are satisfied if
-----------
-2 < x < 1
-----------
The possible values for x that satisfy the inequation are:
-----------
x > 7
OR
-2 < x < 1
-----------
-----------------------------------------------------------
6)a^2+2a<=15
a^2 + 2a <= 15 <==> a^2 + 2a -15 <= 0
The roots of a^2 + 2a -15 are 3.4051 and -4.4051, then:
0 >= a^2 + 2a -15 = (a - 3.4051).(a + 4.4051) ,
then both factors must have DIFFERENT sign, that means:
If (a - 3.4051)>= 0 then must be (a + 4.4051)<= 0
or
a >= 3.4051 AND a <= -4.4051 .
This two situations are incompatible.
The other possibility is when:
(a - 3.4051)<= 0 then must be (a + 4.4051)>= 0
or
a <= 3.4051 AND a >= -4.4051 ,
this situation is possible and we can writte it more clearly as:
----------------------
-4.051 <= a <= 3.4051
----------------------
-----------------------------------------------------------
7)x-x^2<=0
0 >= x - x^2 = x.(1-x)
Then when x >= 0 , must be (1-x)<= 0 or 1 <= x , that is:
0 <= x and 1 <= x,
We have that must be x >= 1 .
When x <= 0 , must be (1-x)>= 0 or 1 >= x , that is:
x <= 0 and x <= 1,
We have that must be x <= 0 .
The possible values for x that satisfy the inequation are:
-------
x >= 1
OR
x <= 0
-------
-----------------------------------------------------------
8)x-3/x+5<0
(x-3)/(x+5) < 0
Obviously x is different from -5 and 3.
If (x+5)< 0, then must be (x-3)> 0 , more clearly:
If x < -5 then x > 3, and this situation is not possible.
If (x+5)> 0, then must be (x-3)< 0 , more clearly:
If x > -5 then x < 3, that means:
-5 < x < 3 .
The possible values for x that satisfy the inequation are:
----------
-5 < x < 3
----------
-----------------------------------------------------------
I hope that this helps you. If you find something unclear or
incomplete, please request for an answer clarification, some sign
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Best regards.
livioflores-ga |
