How do you solve # x^2+8x-2=0# using the quadratic formula?

Answer 1

#- 4 +- 3sqrt2#

Use the improved quadratic formula (Google, Yahoo Search). #D = d^2 = b^2 - 4ac = 64 + 8 = 72# --> #d = +- 6sqrt2# There are 2 real roots: #x = -b/(2a) +- d/(2a) = -8/2 +- (6sqrt2)/2 = - 4 +- 3sqrt2#

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Answer 2

Jameson Allen

To solve the equation (x^2 + 8x - 2 = 0) using the quadratic formula:

Identify the coefficients (a), (b), and (c): (a = 1), (b = 8), and (c = -2).
Apply the quadratic formula: (x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}).
Substitute the values of (a), (b), and (c) into the formula: (x = \frac{{-8 \pm \sqrt{{8^2 - 4 \cdot 1 \cdot (-2)}}}}{{2 \cdot 1}}).
Simplify inside the square root: (x = \frac{{-8 \pm \sqrt{{64 + 8}}}}{{2}}), (x = \frac{{-8 \pm \sqrt{{72}}}}{{2}}).
Simplify further: (x = \frac{{-8 \pm 6\sqrt{2}}}{{2}}).
Divide both terms in the numerator by 2: (x = \frac{{-4 \pm 3\sqrt{2}}}{{1}}).
Therefore, the solutions are: (x = -4 + 3\sqrt{2}) and (x = -4 - 3\sqrt{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.