solve x2 + 2x - 3 < 0 and graph.
Anyone can solve the problem please.
Factor the equation into (x -1)(x + 3) <0 then break it up into two statements x -1<0 or x +3<0 then solve for x<1 or x<-3 then graph that on a number line which is all points less than 1 with an open circle on 1.
v-8<35
To solve the inequality x^2 + 2x - 3 < 0, we need to determine the range of x values that satisfy this inequality.
Step 1: Factor the quadratic equation x^2 + 2x - 3
The factors are obtained by finding two numbers that multiply to -3 and add up to +2. In this case, the factors are (x + 3)(x - 1).
Step 2: Set each factor equal to zero and solve for x.
Setting (x + 3) = 0 gives x = -3, and setting (x - 1) = 0 gives x = 1.
Step 3: Create a number line and plot the values of x on it.
The number line should include the values of x = -3 and x = 1.
Step 4: Pick test values in the intervals created by the values on the number line (e.g., -4, 0, and 2) and evaluate the inequality x^2 + 2x - 3 < 0 for each test value.
• For x = -4, substitute -4 into the inequality: (-4)^2 + 2(-4) - 3 < 0
This simplifies to 16 - 8 - 3 < 0, which is true.
• For x = 0, substitute 0 into the inequality: (0)^2 + 2(0) - 3 < 0
This simplifies to -3 < 0, which is true.
• For x = 2, substitute -2 into the inequality: (2)^2 + 2(2) - 3 < 0
This simplifies to 4 + 4 - 3 < 0, which is false.
Step 5: Analyze the signs of the inequality for each interval.
Based on the results from the test values, we can conclude that the inequality is true for x ∈ (-∞, -3) ∪ (1, ∞) and false for x ∈ [-3, 1].
Step 6: Graph the solution on a graph.
On the coordinate plane, plot a point at x = -3 and another at x = 1. Draw an open (or dashed) circle at each point since the inequality is < (less than). Then, shade the regions to the left of -3 and to the right of 1 to indicate the solution.
Therefore, the solution to the inequality x^2 + 2x - 3 < 0, represented graphically, is:
graph
_________________________
-3 1
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(-∞, -3) (1, ∞)