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

Answer 1

#x = -3#, #x = 1#

Here's how I did it:

To solve for #x#, we factor it. To factor this, you have to find two numbers that multiply up to #-3# and add up to #2#.

These two numbers are #3# and #-1#: #3 * -1 = -3# #3 - 1 = 2#

Now we put the equation like this: #(x+3)(x-1) = 0#

#x+3 = 0 -> x = -3#

#x - 1 = 0 -> x = 1#

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Answer 2

Isaiah Anthony

To solve the quadratic equation (x^2 + 2x - 3 = 0), you can use the quadratic formula. The general form of a quadratic equation is (ax^2 + bx + c = 0), and the quadratic formula is:

[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ]

For the equation (x^2 + 2x - 3 = 0), the coefficients are (a = 1), (b = 2), and (c = -3). Substituting these values into the quadratic formula gives:

[ x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-3)}}{2(1)} ]

[ x = \frac{-2 \pm \sqrt{4 + 12}}{2} ]

[ x = \frac{-2 \pm \sqrt{16}}{2} ]

[ x = \frac{-2 \pm 4}{2} ]

Therefore, the solutions are:

[ x_1 = \frac{-2 + 4}{2} = 1 ]

[ x_2 = \frac{-2 - 4}{2} = -3 ]

So, the solutions to the equation (x^2 + 2x - 3 = 0) are (x = 1) and (x = -3).

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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.