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How do you solve #x^2-x-72=0#?

Answer 1

Genesis Allen

#x=9" "or x=-8# are the two solutions.

To solve a quadratic equation (one with an #x^2# term), there are three options:

In this case the quadratic trinomial can be factored.

#x^2 -x-72 =0" "# find factors of 72 which differ by 1.

#(x-9)(x+8)=0#

Either of the factors could be equal to #0#.

If #x-9=0" "hArr" " x =9#

If #x+8=0" "hArr" "x =-8#

#x=9" "or x=-8# are the two solutions.

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Answer 2

Eva Adamson

To solve the equation (x^2 - x - 72 = 0), you can use the quadratic formula: (x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}), where (a = 1), (b = -1), and (c = -72).

Substitute these values into the formula:

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

Simplify:

(x = \frac{{1 \pm \sqrt{{1 + 288}}}}{2})

(x = \frac{{1 \pm \sqrt{{289}}}}{2})

(x = \frac{{1 \pm 17}}{2})

There are two possible solutions:

(x_1 = \frac{{1 + 17}}{2} = 9)

(x_2 = \frac{{1 - 17}}{2} = -8)

So, the solutions to the equation (x^2 - x - 72 = 0) are (x = 9) and (x = -8).

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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.