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

1 Answer
Douglas K.
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 - 10x + 21 = 0#

We can factor the above:

#(x - 3)(x - 7) = 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#

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