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How do you solve # 2x^2+7x+9=0#?

Answer 1

Genesis Allen

#2x^2+7x+9# is of the form #ax^2+bx+c#, with #a=2#, #b=7# and #c=9#.

The discriminant #Delta# is given by the formula:

#Delta = b^2-4ac = 7^2 - (4xx2xx9) = 49 - 72 = -23#

Since #Delta < 0# the quadratic equation has no real roots. It has two distinct complex roots.

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Answer 2

Genesis Allen

You can solve the quadratic equation (2x^2 + 7x + 9 = 0) using the quadratic formula, which is (x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}), where (a = 2), (b = 7), and (c = 9). Substituting these values into the quadratic formula yields (x = \frac{{-7 \pm \sqrt{{7^2 - 4 \cdot 2 \cdot 9}}}}{{2 \cdot 2}}). Simplify under the square root to get (x = \frac{{-7 \pm \sqrt{{49 - 72}}}}{{4}}), which further simplifies to (x = \frac{{-7 \pm \sqrt{{-23}}}}{{4}}). Since the discriminant ((b^2 - 4ac)) is negative, the solutions are complex. So, the solutions are (x = \frac{{-7 \pm i\sqrt{{23}}}}{{4}}), where (i) is the imaginary unit.

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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.