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How do you find the solution to the quadratic equation #2x^2-5=0#?

Answer 1

Julia Adams

Starting with #2x^2 - 5 = 0#, first add 5 to both sides to get:

#2x^2 = 5#

Divide both sides by 2 to get:

#x^2 = 5/2#

Take the square root of both sides (allowing both positive and negative roots) to get:

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

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Answer 2

Genesis Allen

To find the solution to the quadratic equation (2x^2 - 5 = 0), you can use the quadratic formula:

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

Where (a = 2), (b = 0), and (c = -5). Substituting these values into the formula:

[x = \frac{{-(0) \pm \sqrt{{(0)^2 - 4(2)(-5)}}}}{{2(2)}}]

[x = \frac{{\pm \sqrt{{0 - (-40)}}}}{{4}}]

[x = \frac{{\pm \sqrt{{40}}}}{{4}}]

[x = \frac{{\pm \sqrt{{4 \times 10}}}}{{4}}]

[x = \frac{{\pm 2\sqrt{{10}}}}{{4}}]

[x = \frac{{\pm \sqrt{{10}}}}{{2}}]

So, the solutions to the quadratic equation (2x^2 - 5 = 0) are (x = \frac{{\sqrt{{10}}}}{2}) and (x = -\frac{{\sqrt{{10}}}}{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.