I x + 3 I = -8
I I = absolute value
x=?
thanks!!
Quick, simple math problem?
2008-12-31 3:22 pm
回答 (16)
2008-12-31 3:29 pm
✔ 最佳答案
Case 1:x+3 > 0
|x+3|=x+3
x+3=-8
x=-11
case 2:
x+3 < 0
|x+3| = -(x+3) = -x-3
-x-3 = -8
-x = -5
x=5
x=5
x=-11
2008-12-31 3:26 pm
x is either...
x = -11 or x = -5
x = -11 or x = -5
2008-12-31 4:03 pm
No solution.
Because absolute value can never be equal to a negative number. Don't let the x confuse you. It's just like │-4│ is equal to 4 not -4.
Because absolute value can never be equal to a negative number. Don't let the x confuse you. It's just like │-4│ is equal to 4 not -4.
2008-12-31 3:46 pm
Right I reckon the best way to deal with | | signs is to square both sides and that gets rid of the | |.
So (x+3)^2 = (-8)^2
(x+3)(x+3)= x^2 + 6x + 9 = 64
rearrange to get x^2 + 6x - 55 = 0
Then factorise to get (x+11)(x-5)=0
if x+11=0 then x=-11
if x-5=0 then x=5
there are your answers!
So (x+3)^2 = (-8)^2
(x+3)(x+3)= x^2 + 6x + 9 = 64
rearrange to get x^2 + 6x - 55 = 0
Then factorise to get (x+11)(x-5)=0
if x+11=0 then x=-11
if x-5=0 then x=5
there are your answers!
2008-12-31 3:37 pm
square equation
(x+3)^2 = (-8)^2
expand
x^2+6x+9 = 64
Simplify
x^2+6x-55=0
Solve using quadradic equation
have
a = 1
b = 6
c = -55
x = -6 +/- (squareroot(36+220))/2
hence
-6+16/2 and -6-16/2
x = -11 and/or x= 5
(x+3)^2 = (-8)^2
expand
x^2+6x+9 = 64
Simplify
x^2+6x-55=0
Solve using quadradic equation
have
a = 1
b = 6
c = -55
x = -6 +/- (squareroot(36+220))/2
hence
-6+16/2 and -6-16/2
x = -11 and/or x= 5
參考: brainium
2008-12-31 3:33 pm
there is no answer.. absolute value always come up positive. so if the answers -8 it cant be solved!
2008-12-31 3:30 pm
The key about solving an absolute value problem is to look first at the answer. Because absolute value is a measure of how far away a number is from zero, absolute value must be positive. For example,
l -3 l = 3, and l 3 l = 3. By looking at your problem, you see that the absolute value of x+3 is a negative number. You know that can not be possible, so there is no solution to your question.
l -3 l = 3, and l 3 l = 3. By looking at your problem, you see that the absolute value of x+3 is a negative number. You know that can not be possible, so there is no solution to your question.
2008-12-31 3:29 pm
no soln
2008-12-31 3:28 pm
there is no solution
|x+3| is always positive
|x+3| is always positive
2008-12-31 3:28 pm
The equation is wrong because the absolute value is never negative.
I'll solve assuming it is positive
| x + 3 | = 8
x + 3 = ± 8
x = 3 ± 8 = -5, or 11
I'll solve assuming it is positive
| x + 3 | = 8
x + 3 = ± 8
x = 3 ± 8 = -5, or 11
2008-12-31 3:28 pm
This is a trick question! The absolute value of a number can NEVER be negative. So there is no value for x that can make this equation true! =)
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