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Perform inverse operations until x is isolated.
Treat the inequality as an equation to solve for the boundary points.
Diagram:
Diagram:
By performing inverse operations until x is isolated, we can solve the inequality.
LHS-2≥RHS-2
LHS-x≥RHS-x
.LHS /2.≥.RHS /2.
The solution set to the inequality is x≥ -4. Therefore, we should shade everything that is greater than or equal to -4. Note that this is a non-strict inequality which means we will keep the endpoint closed.
To solve the inequality we must first determine its boundary point(s). To do that, we will treat the inequality as if it was an equation and solve for x.
Inequality:& 2x^2-5x<12
Equality:& 2x^2-5x=12
Let's solve it by using the Quadratic Formula.
LHS-12=RHS-12
Use the Quadratic Formula: a = 2, b= -5, c= -12
- (- a)=a
Calculate power and product
Add terms
Calculate root
State solutions
(I), (II): Add and subtract terms
Calculate quotient
The boundaries of the inequality is x=-1.5 and =4. Let's mark these solutions on a number line. We will keep the points open since this is a strict inequality. To identify where we should shade the number line, we will also include a few test points.
By substituting x with -3, 0, and 5 in the inequality we can determine where the inequality holds true. |c|c|c| [-0.8em] x & 2x^2-5x<12 & Evaluate [0.5em] [-1em] -3 & 2( -3)^2-5( -3)? <12 & 33 ≮ 12 * [0.5em] [-1em] 0 & 2( 0)^2-5( 0)? <12 & 0 < 12 ✓ [0.5em] [-1em] 5 & 2( 5)^2-5( 5)? <12 & 25 ≮ 12 * [0.5em] As we can see, the inequality is true for values between the boundary points. Therefore, we should shade this region.