Inequalities: stuck on how to solve (5+x)(1-2x) ≥0

[M.Adia]

Hi, I have a question I'm stuck on:
(5+x)(1-2x) ≥0

I'm lost on how to solve that.

 

[mmm4444bot]

The quadratic polynomial is already factored, so it's easy to find its roots. You could solve the inequality, by making a sign chart.

Here's a lesson on how to do that. :cool:

 

[HallsofIvy]

It helps to know that "the product of two numbers is positive if and only if both numbers are positive or both numbers are negative".

Here, the inequality is \(\displaystyle (5- x)(1- 2x)\ge 0\) so either
1) \(\displaystyle 5- x\ge 0\) and \(\displaystyle 1- 2x\ge 0\)

or
2) \(\displaystyle 5- x\le 0\) and \(\displaystyle 1- 2x\le 0\).

 

[Deleted member 4993]

Hi, I have a question I'm stuck on:
(5+x)(1-2x) ≥0

I'm lost on how to solve that.

The best way to solve this is through graphing the function y =(5+x)(1-2x)

The domain of the solution would be obvious.

 

[mmm4444bot]

The best way to solve this is through graphing ...

Yes, a picture is worth a thousand math steps!

Yet another way is to use knowledge of parabolas, the vertex formula, sign of leading coefficient, and the roots of the polynomial, and put it all together to reason out the solution.

For example, if you determine that the vertex of some parabola belonging to a function h lies below the x-axis, and you see that the leading coefficient is negative, then no part of the parabola lies above the x-axis. In other words, for that particular example, h(x) is never greater than or equal to zero.

But, for students lacking a lot of practice, this approach would be more difficult than graphing.