Math  /  Algebra

QuestionSolve the inequality: k2>k+6k^{2}>-k+6 Give your answer in interval notation. Enter DNE if there is no solution.

Studdy Solution

STEP 1

What is this asking? Find all the values of kk that make k2k^2 bigger than k+6-k + 6. Watch out! Don't forget to flip the inequality sign when multiplying or dividing by a negative number!
Also, remember to consider *all* possible cases when solving inequalities.

STEP 2

1. Rewrite the inequality
2. Find the roots
3. Analyze the inequality

STEP 3

Let's **rewrite** our inequality to make it easier to work with.
We want everything on one side, so we'll move the k-k and the 66 to the left side.
We do this by *adding* kk to both sides and *subtracting* 66 from both sides: k2>k+6 k^2 > -k + 6 k2+k6>0 k^2 + k - 6 > 0 Now we have a nice **quadratic inequality**!

STEP 4

Now, let's **find the roots** of the quadratic equation k2+k6=0k^2 + k - 6 = 0.
This will help us figure out where the inequality is true.
We can **factor** the quadratic: (k+3)(k2)=0 (k+3)(k-2) = 0 So, our **roots** are k=3k = -3 and k=2k = 2.

STEP 5

Now comes the fun part!
We need to **analyze** the inequality k2+k6>0k^2 + k - 6 > 0.
Think of the quadratic k2+k6k^2 + k - 6 as a **function** of kk.
We know it crosses the kk-axis at 3-3 and 22.
Since the coefficient of the k2k^2 term is **positive** (it's 11), the parabola opens upwards.

STEP 6

This means the quadratic is **positive** when k<3k < -3 and when k>2k > 2.
It's **negative** between 3-3 and 22.
We want the values of kk where the quadratic is **greater than zero**, so our solution is k<3k < -3 or k>2k > 2.

STEP 7

In **interval notation**, the solution is (,3)(2,)(-\infty, -3) \cup (2, \infty).

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