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How do you solve # x²-48=0#?

Answer 1

Eva Adamson

#x = +-4sqrt(3)#

#x^2 - 48 = 0#

#x^2 = 48#

#x = +-sqrt(48)#

#x = +-sqrt(16 *3)#

#x = +-4sqrt(3)#

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Answer 2

Eva Abramson

To solve (x^2 - 48 = 0), you can use the quadratic formula:

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

In this equation, (a = 1), (b = 0), and (c = -48). Substitute these values into the quadratic formula:

[x = \frac{{-0 \pm \sqrt{{0^2 - 4(1)(-48)}}}}{{2(1)}}]

[x = \frac{{\pm \sqrt{{0 + 192}}}}{{2}}]

[x = \frac{{\pm \sqrt{{192}}}}{{2}}]

[x = \frac{{\pm 8\sqrt{{3}}}}{{2}}]

[x = \pm 4\sqrt{{3}}]

So the solutions are (x = 4\sqrt{3}) and (x = -4\sqrt{3}).

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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.