Solve the following quadratic inequality: \(x^2 - x\) - 4 ≤ 2
Solve the following quadratic inequality: \(x^2 - x\) - 4 ≤ 2
- A. \(-3 < x < 2\)
- B. \(-2 ≤ x ≤ 3\)
- C. \(x ≤ -2, x ≤ 3\)
- D. \(-2 < x < 3\)
Correct Answer: B. \(-2 ≤ x ≤ 3\)
Explanation
\(x^2 - x - 4 ≤ 2\)
Subtract two from both sides to rewrite it in the quadratic standard form:
= \(x^2 - x - 4 - 2 ≤ 2 - 2\)
= \(x^2 - x - 6 ≤ 0\)
Now set it = 0 and factor and solve like normal.
= \(x^2 - x\) - 6=0
= \((x - 3)(x + 2)\)=0
\(x\) + 2 = 0 or \(x\) - 3 = 0
\(x\) = -2 or \(x\) = 3
So the two zeros are -2 and 3, and will mark the boundaries of our answer interval. To find out if the interval is between -2 and 3, or on either side, we simply take a test point between -2 and 3 (for instance, \(x\) = 0) and evaluate the original inequality.
= \(x2 - x - 4 ≤ 2\)
= \((0)^2 - (0) - 4 ≤ 2\)
= \(0 - 0 - 4 ≤ 2\)
\(−4 ≤ 2\)
Since the above is a true statement, we know that the solution interval is between -2 and 3, the same region where we picked our test point. Since the original inequality was less than or equal, we include the endpoints.
∴ \(-2 ≤ x ≤ 3.\)