Question 61887: Couls someone help please?
Solve the following inequalitites. Write the answers in interval notation.
x^2 +7x-18=>0
3x-4/2x+1 <=6
Found 3 solutions by jai_kos, ikleyn, MathTherapy:
Answer by jai_kos(139)   (Show Source):
You can put this solution on YOUR website!
Step 1: Write the quadratic inequality in standard form.
This quadratic inequality is already in standard form
Step 2: Solve the quadratic equation,, by factoring to get the boundary point(s).
x^2 +7x-18 =>0
x^2 + 9x -2x -18 = >0
x(x + 9) - 2 (x - 9) => 0
(x -2) (x-9) = > 0
(x -2) = 0 or (x -9) = 0 ---->(1)
x =2 or x = 9
2 and 9 are boundary points.
Step 3:Considering the boudary points, we construct the intervals.
The intervals are (- infinty,2) (2,9) and (9,+infinity)
Step 4: Find the sign of every factor in every interval.
Consider the first interval ( -infinty ,2)
Take a point in the interval, let it be 1.
Now put in equation(1) we get (x - 2) = 1 -2 = -1
(x -9) = 1 -9 = -8
Both the values are negative.
Consider the second interval (2,9),
Take a point in the interval, let it be 3.
Now put in equation(1), we get
(x - 2) = 3 -2 = 1
(x -9) = 3 -9 = -6
One value is positive and the other song is negative.
Consider the Third interval (9, infinity),
Take a point in the interval, let it be 10.
Now put in equation(1), we get
(x - 2) = 10 -2 = 8
(x -9) = 10 -9 = 1
Both the values are positive.
So we consider the interval in which there is variation.
That is,
(2 ,9)
The quadratic equation in the interval notation, we have (2, 9)
2) 3x-4/2x+1 <=6
Mulitply the above equation by (2x +1), we get
3x - 4< = 6 (2x +1)
3x -4 < = 12x + 6
Group all the "x" terms on one side, we get
3x + 12x < = 4 + 6
15x < = 10
Dvide the above equation by 15, we get
x< =10 /15
x <=2/3
Therefore the interval would be (-infinity ,2/3]
Answer by ikleyn(53748)  (Show Source):
You can put this solution on YOUR website! .
Could someone help please?
Solve the following inequalities. Write the answers in interval notation.
(a) x^2 + 7x - 18 => 0
(b) (3x-4)/(2x+1) <= 6
~~~~~~~~~~~~~~~~~~~~~~~
The solutions by @jai_kos are incorrect for both inequalities.
I came to bring correct solutions for both problems.
(a) Your starting inequality is
x^2 + 7x - 18 => 0. (1)
Factor left side quadratic polynomial
(x+9)*(x-2) >= 0.
Critical points are x = -9 and x = 2.
On the left of the critical point x = -9, both factors (x+9) and (x-2) are negative, so their product is positive.
Between the critical points -9 < x < 2, factor (x+9) is positive, while factor (x-2) is negative,
so their product is negative.
On the right of the critical point x = 2, both factors (x+9) and (x-2) are positive, so their product is positive.
Thus the solution set for inequality (1) is x < -9 or x > 2,
or, in interval notation, the union ( , ) U ( , ).
Part (a) is solved correctly.
(b) Your starting inequality is
 <= 6. (1)
Transform it equivalently this way
- 6 <= 0 <<<---=== moving 6 from right side to left side with changing the sign
-  <= 0 <<<---=== writing '6' with the common denominator
 <= 0 <<<---=== simplifying
 <= 0 <<<---=== simplifying further
Now, the left side rational function can be non-positive if and only if
EITHER the numerator is non-negative and denominator is negative
-9x - 10 >= 0 and 2x + 1 < 0 (2)
OR the numerator is non-positive and denominator is positive
-9x - 10 <= 0 and 2x + 1 > 0. (3)
In case (2), 9x <= -10 and x < -1/2, which is the same as
x <= -10/9 and x < -1/2.
These both inequalities, taken together, have the solution set x <= -10/9.
In case (3), 9x >= -10 and 2x > -1, which is the same as
x >= -10/9 and x > -1/2.
These both inequalities, taken together, have the solution set x >= -1/2.
Thus the final solution to the given inequality is this set of real numbers { x <= -10/9 } OR { x >= -1/2 }.
In the interval notation, the solution set is the union of two sets (, ] U ( ,).
Both problems/questions are solved correctly.
Answer by MathTherapy(10806)  (Show Source):
You can put this solution on YOUR website!
Couls someone help please?
Solve the following inequalitites. Write the answers in interval notation.
x^2 +7x-18=>0
3x-4/2x+1 <=6
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Solution for # 2, in INTERVAL NOTATION: ( ] ∪ ( )
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