SOLUTION: -2x^2+7x-10<=(-3)x+2 Possible answers a. (-inf,-3) U (-2,+inf) b. (-inf,2) U (3, inf) c. (-3,-2) d. (2,3)

Algebra ->  Inequalities -> SOLUTION: -2x^2+7x-10<=(-3)x+2 Possible answers a. (-inf,-3) U (-2,+inf) b. (-inf,2) U (3, inf) c. (-3,-2) d. (2,3)      Log On


   



Question 40422: -2x^2+7x-10<=(-3)x+2
Possible answers
a. (-inf,-3) U (-2,+inf)
b. (-inf,2) U (3, inf)
c. (-3,-2)
d. (2,3)

Answer by fazlerabbi(9) About Me  (Show Source):
You can put this solution on YOUR website!
-2%2Ax%5E2%2B7%2Ax-10+%3C=+%28-3%29%2Ax%2B2 given
-2%2Ax%5E2%2B10%2Ax-10+%3C=+2 subtracted (-3)*x from both sides
-x%5E2%2B5%2Ax-5+%3C=+1 multiplied both sides by 1/2
-x%5E2%2B5%2Ax-6+%3C=+0 subtracted 1 from both sides
x%5E2-5%2Ax%2B6+%3E=+0 multiplied both sided by -1
Factoring left side yields
(x-3)(x-2) >= 0
The values of x for which x-2=0 or x-3=0 are x=2 and x=5. These points divide the coordinate line into three intervals,
(-inf,2], (2,3) and [3, +inf)
We need to check points of which of these three intervals give positive sign for the product (x-3(x-2). We shall choose arbitrary points on each of these intervals to determine the sign; these points are called test points. Lets say 1, 2.5 and 4 will be the test points for intervals (-inf,2], (2,3) and [3, +inf) respectively.
For interval (-inf,2] with test point 1 sign of (x-2)(x-3) is (-)(-) = +
For interval (2,3) with test point 2.5 sign of (x-2(x-3) is (+)(-) = -
For interval [3,+inf)with test point 4 sign of (x-2(x-3) is (+)(+) = +
The pattern of signs suggest that the solution set is
(-inf,2] U [3,+inf)