solve (x-4)(x+1) is equal to or less than 28?

Answer 1

distribute the parentheses.

so..
x2 + x - 4x - 4
x2 -3x - 4
solve for x.

then the number will be greater than or less than.

Answer 2

(x-4)(x+1) <= 28
x^2 - 4x + x - 4 <= 28
x^2 - 3x - 32 <= 0

With some help from the quadratic equation, this factors out to
(x - [3+√137]/2) (x - [3-√137]/2) <=0

The values that make this equal to zero are [3+√137]/2 and [3-√137]/2. If the product is less than zero, this means they have to have opposite signs. So either
(x - [3+√137]/2) > 0 and (x - [3-√137]/2) < 0
or
(x - [3+√137]/2) < 0 and (x - [3-√137]/2) > 0

In the first case we have x >[3+√137]/2 and x < [3-√137]/2. This is impossible though, because [3-√137]/2 is smaller than [3+√137]/2. But in the second case we have
x < [3+√137]/2 and x > [3-√137]/2.

So the total solution set is
[3-√137]/2 <= x <= [3+√137]/2

Answer 3

First get everything on one side:

(x - 4)(x + 1) <= 28
x^2 - 3x - 4 <= 28
x^2 - 3x - 32 <= 0

Cannot factor. Use quadratic formula:

(3 +/- √(9 - (4)(1)(-32))) /2

3+/- √137 / 2

Roots: (3+ √137)/2, (3 - √137)/2

(x - (3 + √137)/2)(x - (3 - √137)/2) <= 0

Want product less than zero, so one factor has to be positive and one negative. This happens just in between the two roots.

Solution: [(3 - √137)/2, (3 + √137)/2].

Answer 4

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Answer 5

(x - 4)(x + 1) ≤ 28
x^2 - 3x - 4 ≤ 28
x^2 - 3x ≤ 32
x^2 - 3x + 9/4 ≤ 32 + 9/4
(x - 3/2)^2 ≤ 137/4
x - 3/2 ≤ ± (1/2)√137
3/2 - (1/2)√137 ≤ x ≤ 3/2 + (1/2)√137

Answer 6

(x-4)(x+1) =< 28
x^2-3x -4 =< 28
x^2-3x -32 =<0
x+4.35)(x-7.35) =< 0
So -4.35<= x <= 7.35