Solve the following inequality. Write the solution in interval notation_ x2 3x – 4 > 0
The Correct Answer and Explanation is:
We are given the inequality:x2+3x−4>0x^2 + 3x – 4 > 0x2+3x−4>0
Step 1: Factor the quadratic expression
We need to factor x2+3x−4x^2 + 3x – 4×2+3x−4. We look for two numbers that multiply to −4-4−4 and add up to 333. These numbers are 444 and −1-1−1.
So,x2+3x−4=(x+4)(x−1)x^2 + 3x – 4 = (x + 4)(x – 1)x2+3x−4=(x+4)(x−1)
Now the inequality becomes:(x+4)(x−1)>0(x + 4)(x – 1) > 0(x+4)(x−1)>0
Step 2: Determine the critical points
Set each factor equal to zero:x+4=0⇒x=−4x−1=0⇒x=1x + 4 = 0 \Rightarrow x = -4 \\ x – 1 = 0 \Rightarrow x = 1x+4=0⇒x=−4x−1=0⇒x=1
These are the critical points where the expression changes sign. They divide the number line into three intervals:
- (−∞,−4)(-\infty, -4)(−∞,−4)
- (−4,1)(-4, 1)(−4,1)
- (1,∞)(1, \infty)(1,∞)
Step 3: Test each interval
Pick a test point from each interval and plug it into (x+4)(x−1)(x + 4)(x – 1)(x+4)(x−1):
- In (−∞,−4)(-\infty, -4)(−∞,−4), try x=−5x = -5x=−5:
(x+4)(x−1)=(−5+4)(−5−1)=(−1)(−6)=6>0(x + 4)(x – 1) = (-5 + 4)(-5 – 1) = (-1)(-6) = 6 > 0(x+4)(x−1)=(−5+4)(−5−1)=(−1)(−6)=6>0 - In (−4,1)(-4, 1)(−4,1), try x=0x = 0x=0:
(x+4)(x−1)=(0+4)(0−1)=(4)(−1)=−4<0(x + 4)(x – 1) = (0 + 4)(0 – 1) = (4)(-1) = -4 < 0(x+4)(x−1)=(0+4)(0−1)=(4)(−1)=−4<0 - In (1,∞)(1, \infty)(1,∞), try x=2x = 2x=2:
(x+4)(x−1)=(2+4)(2−1)=(6)(1)=6>0(x + 4)(x – 1) = (2 + 4)(2 – 1) = (6)(1) = 6 > 0(x+4)(x−1)=(2+4)(2−1)=(6)(1)=6>0
Step 4: Include correct intervals
We want the product to be greater than zero, so we take the intervals where the product is positive. That is:
- (−∞,−4)(-\infty, -4)(−∞,−4)
- (1,∞)(1, \infty)(1,∞)
We do not include the endpoints −4-4−4 and 111 because the inequality is strictly greater than zero (not greater than or equal to zero).
Final Answer:
(−∞,−4)∪(1,∞)\boxed{(-\infty, -4) \cup (1, \infty)}(−∞,−4)∪(1,∞)
Explanation
Solving a quadratic inequality like x2+3x−4>0x^2 + 3x – 4 > 0x2+3x−4>0 involves a few important steps. First, we factor the quadratic expression. Factoring means rewriting the expression as a product of two binomials. In this case, the expression factors as (x+4)(x−1)(x + 4)(x – 1)(x+4)(x−1). This gives us a product of two linear expressions, and we now want to find when this product is greater than zero.
Next, we identify the critical points. These are the values of xxx that make the expression equal to zero. Setting each factor equal to zero gives x=−4x = -4x=−4 and x=1x = 1x=1. These values split the number line into three regions. Each region needs to be tested to determine whether the inequality holds there.
We pick one test point in each region and plug it into the factored form (x+4)(x−1)(x + 4)(x – 1)(x+4)(x−1). If the result is positive, the inequality is satisfied in that region. If it is negative, it is not. For x=−5x = -5x=−5 and x=2x = 2x=2, the product is positive. For x=0x = 0x=0, the product is negative.
Finally, we include only the intervals where the expression is positive. Since the inequality is strict (uses > and not ≥), we do not include the boundary points x=−4x = -4x=−4 or x=1x = 1x=1.
Therefore, the solution in interval notation is:(−∞,−4)∪(1,∞)(-\infty, -4) \cup (1, \infty)(−∞,−4)∪(1,∞)
This shows all values of xxx where the original inequality is true.
