Solve (x-1)(x-4)leq0quad12

$$( x - 1 ) ( x - 4 ) \leq 0 \quad 12$$

Solve for x

$$x\in \begin{bmatrix}1,4\end{bmatrix}$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x-1 by x-4 and combine like terms.

$$x^{2}-5x+4\leq 0\times 12$$

Multiply 0 and 12 to get 0.

$$x^{2}-5x+4\leq 0$$

To solve the inequality, factor the left hand side. Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$x^{2}-5x+4=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, -5 for b, and 4 for c in the quadratic formula.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 1\times 4}}{2}$$

Do the calculations.

$$x=\frac{5±3}{2}$$

Solve the equation x=\frac{5±3}{2} when ± is plus and when ± is minus.

$$x=4 x=1$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-4\right)\left(x-1\right)\leq 0$$

For the product to be ≤0, one of the values x-4 and x-1 has to be ≥0 and the other has to be ≤0. Consider the case when x-4\geq 0 and x-1\leq 0.

$$x-4\geq 0 x-1\leq 0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-4\leq 0 and x-1\geq 0.

$$x-1\geq 0 x-4\leq 0$$

The solution satisfying both inequalities is x\in \left[1,4\right].

$$x\in \begin{bmatrix}1,4\end{bmatrix}$$

The final solution is the union of the obtained solutions.

$$x\in \begin{bmatrix}1,4\end{bmatrix}$$