How do you solve the inequality #3

Question

How do you solve the inequality #3 <= x^2 - 8x + 15#?

Nathan Axel · Accepted Answer

Because the coefficient of the #x^2# term is greater than 0, we know that the domain between the roots will cause the quadratic to be less than 0. We shall include all values of x except this region.

Given: #3 <= x^2 - 8x + 15# Subtract 3 from both sides: #0 <= x^2 - 8x + 12# Flip the inequality: #x^2 - 8x + 12 >= 0# Because the coefficient of the #x^2# term is greater than 0, we know that the quadratic represents a parabola that opens upward. Therefore, the quadratic will be less than 0 between the two roots and greater than or equal to 0 elsewhere. Let's find the roots by factoring: #(x-2)(x-6) = 0# #x = 2 and x = 6# Therefore, we know that the two domains that will cause the quadratic to be greater than or equal to 0 are: #x <= 2# and #x >=6# Here is a graph to prove it: graph{x^2-8x+12 [-10, 10, -5, 5]}

Julia Adams · Answer

undefined To solve the inequality $3 \leq x^2 - 8x + 15$:

1. Move all terms to one side of the inequality: $x^2 - 8x + 15 - 3 \geq 0$.
2. Simplify the expression: $x^2 - 8x + 12 \geq 0$.
3. Factor the quadratic expression: $(x - 6)(x - 2) \geq 0$.
4. Find the critical points by setting each factor equal to zero: $x - 6 = 0 $ and $x - 2 = 0$.
5. Solve for $x$: $x = 6$ and $x = 2$.
6. Plot these critical points on a number line.
7. Test each interval to determine where the inequality holds true.
8. Identify the intervals where the inequality is satisfied.

The solution to the inequality $3 \leq x^2 - 8x + 15$ is $x \in [2, 6]$.

How do you solve the inequality #3 &lt;= x^2 - 8x + 15#?

How do you solve the inequality #3 <= x^2 - 8x + 15#?