solve for the variable:
6x2 - 6x - 36 = 0
(6x2 means 6x squared)
solve the following inequalities:
-6x<18
-16x+4<-8x
factor by undoing foil:
x2+5x+6
(x2 means x squared)
x2-5x-14
2006-09-01 04:59:35 · 10 answers · asked by nicknick 1 in Science & Mathematics ➔ Mathematics
I don't think I'm refusing to believe some lazy people use Y!Answers to have others make their homework anymore.
You, if you post some problem on the Maths section, at least make it a challenge...
2006-09-01 05:05:30 · answer #1 · answered by Kazeed 2 · 0⤊ 0⤋
You can easily solve this equation by factoring:
Your equation is:
6x^2 - 6x - 36 = 0
Divide all the terms by 6, giving you:
6{x^2 - 6x - 6} = 0 Now factor the equation in the brackets by the FOIL Method: (You cannot discard the 6 that you factored out of the original equation, that is why it is in front of the brackets!)
You factor the x^2 -6x - 36, resulting in:
(x + 2) (x - 3) (Place these back in the brackets so you don't forget how the statement is written)
6{(x+2) (x - 3)} = 0
Set each of the terms in the brackets equal to zero:
x + 2 = 0 x = -2
x - 3 = 0 x = 3
Now you have to go back and plug each answer into the original equation, to see if both answers are correct, or if only one of them fits.
6(-2)^2 -6(-2) - 36 = 0 This is a false statement
x is not = -2
6(+3)^2 -6(+3) - 36 = 0 This is a true statement
x = 3
The second problem can be solved by treating the statement like an equation:
- 6x < 18 Divide both sides by -6 - 6x/ -6 < 18/ -6
x < -3 This tells you that the value of x must be a negative number, and because it is LESS than 18, it can't equal -3. Now you have to make your conditions:
{x l x < - 3, x E R}
Where E stands for element and R stands for the set of Real numbers
The next one is a little more complicated:
-16x + 4 < -8x You want to add 16x to both sides of the inequality, resulting in:
-16x + 4 + 16x < -8x + 16x Combine like terms and get:
+ 4 < 8x Divide both sides by 8, giving you:
+4/8 < 8x/8
x < 1/2 Your conditions are:
{x l x< 1/2, x E R }
Your last problem involves factoring by the foil method:
x^2 - 5x - 14
(x + 2) ( x - 7) Since there is no coefficient for the x^2 term, the first term in each parentheses will be an x. Since the last term of the problem is -14, the only combination that will give me a 5 in the middle term will be 2 x 7.
2006-09-01 07:22:14 · answer #2 · answered by nammy_410 2 · 0⤊ 0⤋
first you have
6x^2 - 6x - 36 = 0 divide all terms by 6 first
x^2 -x -6 = 0 then factor and you get
(x-3)(x+2) = 0 so
x = 3 or -2
-6x < 18 divide through by -6 and when you divide by a negative the sign swithes
x > 3
-16x + 4 < -8x first group the terms
-8x < -4 then divide by -8
x>1/2
x^2 +5x +6 = 0 first factor to get
(x+2)(x+3) = 0 so
x = -3 or -2
x^2 -5x -14 = 0 first you have to factor
(x-7)(x+2) = 0 so
x = 7 or -2
2006-09-01 05:18:33 · answer #3 · answered by tlets 2 · 0⤊ 0⤋
6x² - 6x - 36 = 0
6(x² - x - 6) = 0
Factor the 6 out The equation becomes 6(x² - x -6) = 0
6( x + 7)(x - 3
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
-6x < 18
-6x/-6 > 18/-6
x = - 3
When dividing the inequality sign changes from < to >
- - - - - - - - - - - - - - - - - - - - - - - - -
-16x + 4 < - 8x
-4 -4
Subtracting -4 from both sides
-16x < -8x - 4
+8x +8x
Adding + 8x to both sides
-8x < -4
-8x/-8 > -4/-8
Dividing both sides by - 8. When dividing the inequality sign changes.
x = 1/2
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
x² + 5x + 6
(x + 2)(x + 3)
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
x² - 5x - 14
(x - 7)(x + 2)
2006-09-01 06:48:06 · answer #4 · answered by SAMUEL D 7 · 0⤊ 0⤋
12
2006-09-01 05:02:59 · answer #5 · answered by 34 RIP 3 · 0⤊ 1⤋
6x2 - 6x - 36 = 0
6(x^2 - x - 6) =0
6(x^2 -3x +2x -6) = 0
6 { x(x-3) + 2 (x - 3) } =0
6 (x+2) (x -3) =0
Therefor x= -2 or x= 3..............i
-6x< 18
By changing the sign from negative to positive chage less to grater
6x > -18
Divide by 6
x>-6....................................ii
-16x+4<-8x
Add 16x both sides
4<8x
Divide by 8
o.5
x2+3x+2x +6
x(x+3) + 2 (x+3)
(x+3)(x+2).............................iv
x2-5x-14
x2-7x + 2x -14
x(x - 7) + 2 ( x-7)
(x-7) (x+2).............................v
2006-09-01 05:43:48 · answer #6 · answered by Amar Soni 7 · 0⤊ 0⤋
the perfect answer demands which you persist with the regulations of mathematical operations. First, you will possibly desire to upload what's interior the parenthesis, then multiply that by ability of what's outdoors, accompanied by ability of branch. So 6/2(a million+2) = 6/2(3) = 6/6 = a million
2016-11-06 05:28:32 · answer #7 · answered by ? 4 · 0⤊ 0⤋
The challenge does not exist, and what is your point?
2006-09-01 05:06:24 · answer #8 · answered by Anonymous · 0⤊ 0⤋
x=3
x<3
unsolvable (I think)
You got me.
You got me again.
2006-09-01 05:14:46 · answer #9 · answered by professionaleccentric 5 · 0⤊ 0⤋
x=3,-2
6(-2)^2-6(-2)-36=0
24+12-36=0
6(3)^2-6(3)-36=0
54-18-36=0
x>-3
-6(-3)=18
-6(-2)<18
-6(-4)>18
x>(1/2)
-16(1/2)+4<-8(1/2)
-4=-4
-16(1)+4<-8(1)
-12<-8
-16(0)+4>-8(0)
4>0
(x+2)(x+3)
(x-7)(x+2)
2006-09-01 05:11:58 · answer #10 · answered by Scott L. 2 · 0⤊ 0⤋