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How do you find all the zeros of #x^2 + x -72 = 0#?

Answer 1

Ariana Alvarez

#x=-9 or x=8#

#x^2+x-72=0# #(x+9)(x-8)=0# #x=-9 or x=8#

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Answer 2

Lily Anderson

To find the zeros of (x^2 + x - 72 = 0), you can use the quadratic formula, which states that for any quadratic equation in the form (ax^2 + bx + c = 0), the solutions for (x) are given by:

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

For the given equation (x^2 + x - 72 = 0), the coefficients are (a = 1), (b = 1), and (c = -72). Substituting these values into the quadratic formula:

[x = \frac{{-1 \pm \sqrt{{1^2 - 4(1)(-72)}}}}{{2(1)}}]

[x = \frac{{-1 \pm \sqrt{{1 + 288}}}}{2}]

[x = \frac{{-1 \pm \sqrt{{289}}}}{2}]

[x = \frac{{-1 \pm 17}}{2}]

So, the solutions for (x) are (x = \frac{{-1 + 17}}{2}) and (x = \frac{{-1 - 17}}{2}), which simplify to (x = 8) and (x = -9), respectively. Therefore, the zeros of the equation are (x = 8) and (x = -9).

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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.