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How do you solve #x^2 + 5x + 6 = 0# algebraically?

Answer 1

Eva Adamson

#x = -3# OR #x = -2#

We have: #x^2 +5x+6=0#

Step 1: Factor Since we have already set the quadratic equal to 0, we can just factor. #(x+3)(x+2) = 0#

Step 2: Set each factor equal to 0 #x+3 = 0# OR #x+2 = 0# #x = -3# OR #x = -2#

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Answer 2

Eva Adamson

To solve the quadratic equation (x^2 + 5x + 6 = 0) algebraically, you can use the quadratic formula, which is (x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}), where (a), (b), and (c) are the coefficients of the quadratic equation (ax^2 + bx + c = 0).

For the given equation (x^2 + 5x + 6 = 0), (a = 1), (b = 5), and (c = 6).

Substitute these values into the quadratic formula:

(x = \frac{{-5 \pm \sqrt{{5^2 - 4(1)(6)}}}}{{2(1)}})

Solve the expression under the square root:

(x = \frac{{-5 \pm \sqrt{{25 - 24}}}}{2})

(x = \frac{{-5 \pm \sqrt{1}}}{2})

(x = \frac{{-5 \pm 1}}{2})

This gives two possible solutions:

(x = \frac{{-5 + 1}}{2} = \frac{-4}{2} = -2)
(x = \frac{{-5 - 1}}{2} = \frac{-6}{2} = -3)

So, the solutions to the equation (x^2 + 5x + 6 = 0) are (x = -2) 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.