How do you solve #x^2 +11x+30=0#?

Answer 1

# x = -6 or x= -5#

First, we must factorize. Locate factors of 30 that sum up to 11. The signs in the brackets will always be positive.

#5xx6 = 30, and 5 + 6 = 11#

#(x+5)(x+6) = 0

Since multiplying two factors yields a result of 0, one of the factors must be 0.

#if x + 5 = 0, x = -5# #if x +6 = 0, x =-6#
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Answer 2

To solve the quadratic equation (x^2 + 11x + 30 = 0), you can use the factoring method or the quadratic formula. Factoring involves finding two numbers that multiply to give the constant term (30) and add to give the coefficient of the linear term (11). In this case, those numbers are 5 and 6. So, you can factor the equation as ((x + 5)(x + 6) = 0). Setting each factor equal to zero gives you two equations: (x + 5 = 0) and (x + 6 = 0). Solving these equations gives you (x = -5) and (x = -6), which are the solutions to the quadratic equation. Alternatively, you can use the quadratic formula, (x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}), where (a = 1), (b = 11), and (c = 30). Substituting these values into the formula and solving gives you the same solutions, (x = -5) and (x = -6).

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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.

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.

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.

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.

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