Chapter A, Problem 24E

To determine

To solve the below inequality in terms of intervals and illustrate the solution set on the real number line -

  $x^3+3x<4x^2$

Answer to Problem 24E

The solution of the inequality is $x<0$ or $1<x<3$ and the solution set on a real number line -

  

Explanation of Solution

Given: Inequality: $x^3+3x<4x^2$

Formula Used:

An inequality compares two values, showing if one is less than, greater than, or simply not equal to another value.

Real number line is the line whose points are the real numbers. 

Calculation:

Given: Inequality equation is $x^3+3x<4x^2$

Simplifying the above inequality, we have:

  $x^3+3x<4x^2$

  $\begin{array}{l}{x}^3-4x^2+3x<0\\ x\left(x^2-4x+3\right)<0\\ x\left(x-1\right)\left(x-3\right)<0\end{array}$

To solve the above inequalities, we need to find the different intervals for which the inequality gives a value greater than $0$ .

When $x<0$ :

  $x$ is negative, $\left(x-1\right)$ is negativeand $\left(x-3\right)$ is negative.

Thus, $x\left(x-1\right)\left(x-3\right)<0$

So, $x<0$ isone of the solutions.

When $0<x<1$ :

  $x$ is positive, $\left(x-1\right)$ is negativeand $\left(x-3\right)$ is negative.

Thus, $x\left(x-1\right)\left(x-3\right)>0$

So, $0<x<1$ is not one of the solutions.

When $1<x<3$ :

  $x$ is positive, $\left(x-1\right)$ is positive and $\left(x-3\right)$ is negative.

Thus, $x(x+1)(x-1)<0$

So, $1<x<3$ is one of the solutions.

When $x>3$ :

  $x$ is positive, $\left(x-1\right)$ is positive and $\left(x-3\right)$ is positive.

Thus, $x\left(x-1\right)\left(x-3\right)>0$

So, $x>3$ is not one of the solutions.

Combining all the solutions, we have the solution set as:

  $x<0$ or $1<x<3$

Drawing the above inequality on a real number line, we have:

  

Conclusion:

Hence, the solution of the inequality is $x<0$ or $1<x<3$ and the solution set on a real number line -