Question

How do you solve the inequality `3x^2<=11x+4` and write your answer in interval notation?

Answer 1

The solution is `x in [-1/3, 4]`

Explanation:

Let's rewrite and factorise the inequality

`3x^2<=11x+4`

`3x^2-11x-4<=0`

`(3x+1)(x-4)<=0`

Let `f(x)=(3x+1)(x-4)`

Now we can build the sign chart

`color(white)(aaaa)` `x` `color(white)(aaaaa)` `-oo` `color(white)(aaaa)` `-1/3` `color(white)(aaaa)` `4` `color(white)(aaaaa)` `+oo`

`color(white)(aaaa)` `3x+1` `color(white)(aaaaa)` `-` `color(white)(aaaaa)` `+` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `x-4` `color(white)(aaaaaa)` `-` `color(white)(aaaaa)` `-` `color(white)(aaaa)` `+`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaaa)` `+` `color(white)(aaaaa)` `-` `color(white)(aaaa)` `+`

Therefore,

`f(x)<=0` when `x in [-1/3, 4]`

Answer 2

`[-1/3,4]`

Explanation:

One way is to get a zero (0) on one side of the inequality.

`rArr3x^2-11x-4<=0`

factorising the quadratic gives.

`(3x+1)(x-4)<=0`

The zeroes of the quadratic are `x=-1/3" and "x=4`

These values will be part of the solution since we have an 'or equal to' in the inequality.

We now require to consider the 'less than' part of the inequality.

For the quadratic to be less than zero.

`• 3x+1<0color(blue)" and "x-4>0`

`rArrx<-1/3color(blue)" and "x>4to" impossible"`

`color(red)"OR"`

`• 3x+1>0color(blue)" and " x-4<0`

`rArrx> -1/3color(blue)" and " x<4`

`rArr-1/3<=x<=4" is the solution"`

`"in interval notation" [-1/3,4]`