How to solve (x)^(2)-6x+5 >= 0

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

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Question

$${ x }^{ 2 } -6x+5 \geq 0$$

Solve for x

$x\in (-\infty,1]\cup [5,\infty)$

Steps Using the Quadratic Formula

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 $ax^{2}+bx+c=0$.

$$x^{2}-6x+5=0$$

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 $a$, $-6$ for $b$, and $5$ for $c$ in the quadratic formula.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 1\times 5}}{2}$$

Do the calculations.

$$x=\frac{6±4}{2}$$

Solve the equation $x=\frac{6±4}{2}$ when $±$ is plus and when $±$ is minus.

$$x=5$$ $$x=1$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-5\right)\left(x-1\right)\geq 0$$

For the product to be $≥0$, $x-5$ and $x-1$ have to be both $≤0$ or both $≥0$. Consider the case when $x-5$ and $x-1$ are both $≤0$.

$$x-5\leq 0$$ $$x-1\leq 0$$

The solution satisfying both inequalities is $x\leq 1$.

$$x\leq 1$$

Consider the case when $x-5$ and $x-1$ are both $≥0$.

$$x-1\geq 0$$ $$x-5\geq 0$$

The solution satisfying both inequalities is $x\geq 5$.

$$x\geq 5$$

The final solution is the union of the obtained solutions.

$$x\leq 1\text{; }x\geq 5$$

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