Solve this inequality:
(6-5x)^2 is less than or equal to -4
2006-10-18 21:42:36 · 11 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Q. (6-5x)^2<=-4
Ans. 36-60x+25x^2<=-4;
25x^2-60x+40<=0;
dividing by 5;
5x^2-12x+8<=0;
subtract 1 from both sides;
5x^2-12x+8-1<=0-1;
5x^2-12x+7<=-1;
5x^2-5x-7x+7<=-1;
5x(x-1)-7(x-1)<=-1;
(x-1)(5x-7)<=-1;
either
(x-1)<=-1 or (5x-7)<=-1
x<= 0 x<= 6/5
on checking it concludes that your question is wrong
2006-10-18 22:11:35 · answer #1 · answered by Anonymous · 0⤊ 0⤋
(6-5x)^2 <= -4
A square of a number is always positive (or zero). Thus, it will never be less than or equal to -4.
By "close inspection", it can be found out that the inequality has no solutions; the set of solutions is empty.
2006-10-19 05:21:45 · answer #2 · answered by realdan 3 · 0⤊ 0⤋
Impossible my dear sir.
But if we are working with the imaginary world, there is an answer:
6-5x<=2i
or
6-5x>=2i
Let's solve as an equality:
Than we have two possible answers:
x>=(6-2i)/5
x<=(6-2i)/5
Note: i is an imaginary number which represents the square root of -1
x<=(6-2i)/5
2006-10-19 05:35:33 · answer #3 · answered by Arc 2 · 1⤊ 0⤋
(6-5x)^2=4
6-5x = ±2
say +2 ;
5x=6-2
x=4/5
say -2
5x = 6+2
5x = 8
x = 5/8
2006-10-19 05:02:04 · answer #4 · answered by caveman0101 1 · 0⤊ 1⤋
anything squared is always positive unless you are talking about an exotic
algebraic operation.
I did notice the other (first) answer found a real solution to the SQRT of a negative number.
2006-10-19 05:06:37 · answer #5 · answered by sydney m 2 · 2⤊ 0⤋
x is greater than or equal to <1.1314>
(6-5x)^2 ⤠-4
-5x ⤠2i -6
5x/5 >= (-2i + 6)/5
x >= -0.4 i + 1.2
x^2 >= -0.16 + 1.44
then x >= 1.1314
2006-10-19 05:45:45 · answer #6 · answered by bada_ping 2 · 0⤊ 1⤋
Look at the problem: the left side is a square, so it is > 0, the right side is negative... there is no (real) solution.
2006-10-19 04:50:25 · answer #7 · answered by sofarsogood 5 · 2⤊ 0⤋
36-60x+25x^2 ⤠-4
25x^2 -60x + 40 ⤠0
5x^2 -12 x + 8 ⤠0
x ⤠(12 +/- sqrt(144-160))/10
x ⤠(12 +/- sqrt(-16))/10
x ⤠(12 +/- j4)/10 = 1.2 +/- j0.4
1.2 - j0.4 ⤠x ⤠1.2 + j0.4, or
x ⤠1.2649 @ ±18.435°
2006-10-19 05:39:19 · answer #8 · answered by Helmut 7 · 0⤊ 1⤋
This does not have a solution in real domain. as LHS is a positive number
2006-10-19 05:20:23 · answer #9 · answered by Mein Hoon Na 7 · 1⤊ 0⤋
x = 1.6i
Working
-4 ^ (0.5) = -2i
-5x = (-2 - 6)i
5x = 8i
x = 1.6i
Glad I could do your homework for you.
_
2006-10-19 04:44:30 · answer #10 · answered by GoogleRules 3 · 0⤊ 2⤋