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How do you solve #x^2+2x-3>0#?

Answer 1

Isaiah Anthony

You first find the zero-points, and see what happens in-between

#->(x-1)(x+3)=0->x=1orx=-3#

For any #x#-value in between we get a negative #y#, so the answer must be: #x<-3orx>1# graph{x^2+2x-3 [-10, 10, -5, 5]}

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Answer 2

Eva Abramson

To solve the inequality \( x^2 + 2x - 3 > 0 \): 1. Factor the quadratic expression: \( (x + 3)(x - 1) > 0 \). 2. Identify the critical points by setting each factor equal to zero: \( x + 3 = 0 \) and \( x - 1 = 0 \). 3. Solve for \( x \) to find the critical points: \( x = -3 \) and \( x = 1 \). 4. Plot these critical points on a number line. 5. Test a value in each interval to determine the sign of the expression. 6. Since the inequality is greater than zero, we're looking for where the expression is positive. 7. The solution is \( x < -3 \) or \( x > 1 \).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How do you solve #x^2+2x-3&gt;0#?

How do you solve #x^2+2x-3>0#?