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Math inequality problem

Can anyone please help me solve this problem:

Find the values of x for which x^2(x−4)>(x−12)
Put everything onto one side so you have a cubic on one side and 0 on the other. Then try if you can factorise the cubic and draw a graph of it. Shade in the areas that satisfy the inequality.
(edited 4 years ago)
X²(x-4)>(x-12)
If we are to find x, expand the left hand equation then collect like terms
This will be:
X³-4x²>X-12
Rewriting gives
X³-4x²-X+12>0
This is a cubic function. So we solve it this way
X² -X-4
X³-4x²-X+12
X-3. X³-3x²
-X²-X+12
-X²+3X
-4X+12
-4X+12
=0
So we solve
X³-4x²-X+12 using long division method, first we find the factors of the equation by substituting values, the one that gives you zero is a factor of X³-4x²-X+12
And I found that X=3 is a factor so X-3=0 is a factor of the equation.
We use long division to factorize the equation as show above.
So the equation can be written as X-3(X²-X-4)=0
So here we can introduce our inequality.
X-3(X²-X-4)>0
We solve the quadratic equation in bracket first using the quadratic formula you will have two values of x
So the final answer will have 3 values of x. Since it is a cubic function.
X>3, and the other 2 values of x you will get after solving the quadratic equation X²-X-4 let's say first value will be a, and second value be b. The answer will be
X>3
X>a
X>b
Thanks for reading



Collecting like terms
Reply 3
Avatar for mqb2766
mqb2766
21
Original post by BreenaBEE
X²(x-4)>(x-12)
If we are to find x, expand the left hand equation then collect like terms
This will be:
X³-4x²>X-12
Rewriting gives
X³-4x²-X+12>0
This is a cubic function. So we solve it this way
X² -X-4
X³-4x²-X+12
X-3. X³-3x²
-X²-X+12
-X²+3X
-4X+12
-4X+12
=0
So we solve
X³-4x²-X+12 using long division method, first we find the factors of the equation by substituting values, the one that gives you zero is a factor of X³-4x²-X+12
And I found that X=3 is a factor so X-3=0 is a factor of the equation.
We use long division to factorize the equation as show above.
So the equation can be written as X-3(X²-X-4)=0
So here we can introduce our inequality.
X-3(X²-X-4)>0
We solve the quadratic equation in bracket first using the quadratic formula you will have two values of x
So the final answer will have 3 values of x. Since it is a cubic function.
X>3, and the other 2 values of x you will get after solving the quadratic equation X²-X-4 let's say first value will be a, and second value be b. The answer will be
X>3
X>a
X>b
Thanks for reading



Collecting like terms

The aim is to give hints, not worked solutions.
Also, the last set of inequalities are not correct.
The values of a and be you find them from the quadratic equation they represent x
Reply 5
Avatar for mqb2766
mqb2766
21
Original post by BreenaBEE
The values of a and be you find them from the quadratic equation they represent x

Sure, but the answer is not
x>a
X>b
X>3
Correct where there is a mistake please. That's why we are here to learn
Original post by BreenaBEE
Correct where there is a mistake please. That's why we are here to learn


Towards the end.

You have (x−3)(x2−x−4)>0(x-3)(x^2-x-4) > 0, which is a product of two expressions being greater than zero.

A product is greater than zero ONLY when both expressions are *either* greater than zero, or less than zero. These are your two cases. Mathematically, this means;

Case 1: x−3>0x-3 > 0 and x2−x−4>0x^2 - x - 4 > 0

Case 2: x−3<0x-3 < 0 and x2−x−4<0x^2 - x - 4 < 0.


So you need to solve both cases for a range of values of xx satisfying them, then join them together and that's the overall range of values for which our cubic is greater than zero.

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