I need help with solving this maths question.....with workings :)?

Question

(x-4)(x-1)<=28

Answer 1

Multiply out:
x^2 - 5x + 4 <= 28
x^2 - 5x - 24 <= 0
Factorise:
(x - 8)(x + 3) <= 0
(x - 8)(x + 3) is negative or zero.

To get a negative or zero product, once factor must be positive or zero, and the other negative or zero.

Consider first the case where x + 3 >= 0.
Then:
x - 8 <= 0
x <= 8.
The condition x + 3 >= 0 gives x >= -3.
Combining these results:
-3 <= x <= 8.

Now consider the case where x + 3 <= 0.
Then:
x - 8 >= 0
x >= 8.
x + 3 <= 0 gives x <= -3.
Clearly there are no solutions from this option, as x <= -3 cannot be combined with x >= 8.

The answer is therefore:
-3 <= x <= 8.

Answer 2

Q:(x-4)(x-1)<=28
first,
open the braqcuets:
x^2 - 4x - x + 4<=28
take 28 to the other side
x^2 - 5x - 24<=0
factorise this expression:
x^2 - 8x+ 3x- 24<=0
hence,
(x-8)(x+3)<=0
now,
use the rule of inequalities:
1)if(x-a)(x-b)<=0 and a a<=xand, x<=b
2)if(x-a)(x-b)>=0 and a>b;
a>=x and b<=x

as you can clearly see , rule 1 applies to your question
hence,
x>-3 and x<8
{-3 because equating x+3 =0 we get x=-3 which less than 8 from x-8=0. rest you can infer from the first law)

Answer 3

Ok, first you have to make this a solvable equation (polynomial):

(x - 4)(x - 1) <= 28

=

x^2 - 5x + 4 <= 28

=

x^2 - 5x - 24 = 0

From here, you can factor this polynomial:

(x + 3)(x - 8) = 0

So, (because any number mult. by 0 = 0):

x = -3 or x = 8

In this case both answers check, so your number set (answer) would be {-3,8}

Hope this helps!

Answer 4

(x-4)(x-1)<=28
x^2 - 4x - x + 4 <= 28
x^2 - 5x + 4 <= 28
x^2 - 5x + 4 - 28 <= 0
x^2 - 5x - 24 <= 0
x^2 - 8x + 3x - 24 <= 0
(x-8)(x+3) <= 0
-> x-axis intercepts at x=8, x=-3.
-> positive quadratic: negative between intercepts.
-> -3 <= x <= 8.

Answer 5

x^2 - 5x + 4 - 28 <= 0
x^2 - 5x - 24 <= 0
(x-8)(x+3) <= 0