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How do you solve #x^2+3x+1=0#?

Answer 1

Abigail Allen

We can solve this by completing the square ...

#0 = x^2+3x+1#

#= x^2+3x+9/4 - 9/4 + 1#

#=(x+3/2)^2-5/4#

Add #5/4# to both ends to get:

#(x+3/2)^2 = 5/4#

#x+3/2 = +-sqrt(5/4) = +-sqrt(5)/sqrt(4) = +-sqrt(5)/2#

Subtract #3/2# from both ends to get:

#x = -3/2 +- sqrt(5)/2#

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Answer 2

Isaiah Anthony

To solve the quadratic equation (x^2 + 3x + 1 = 0), you can use the quadratic formula, which states that for an equation of the form (ax^2 + bx + c = 0), the solutions are given by:

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

In this case, (a = 1), (b = 3), and (c = 1). Substituting these values into the formula:

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

[x = \frac{{-3 \pm \sqrt{{9 - 4}}}}{{2}}]

[x = \frac{{-3 \pm \sqrt{5}}}{{2}}]

So, the solutions are:

[x = \frac{{-3 + \sqrt{5}}}{{2}}] and [x = \frac{{-3 - \sqrt{5}}}{{2}}]

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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.