# How do you solve x^2-10x+21<=0?

Oct 12, 2016

$3 \le x \le 7$

#### Explanation:

Please observe the coefficient for the ${x}^{2}$ term is positive; this means that the parabola opens up and, if the quadratic has roots, the quadratic will be less than between the two roots.

Let's see if the quadratic has roots:

${x}^{2} - 10 x + 21 = 0$

We can factor the above:

$\left(x - 3\right) \left(x - 7\right) = 0$

The quadratic crosses the x axis at $x = 3$ and $x = 7$. Between these two x values, the quadratic is less than or equal to zero. Therefore, the answer is:

$3 \le x \le 7$