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How do you solve #0=p^2-p-11#?

Answer 1

Julia Adams

Use the quadratic formula to find #p = (1+-3sqrt(5))/2#

If #ap^2+bp+c = 0# then

#p = (-b+-sqrt(b^2-4ac))/(2a)#

We have #a=1#, #b=-1#, #c=-11#

So:

#p = (1+-sqrt(1+44))/2 = (1+-sqrt(45))/2#

#= (1+-sqrt(3^2*5))/2 = (1+-3sqrt(5))/2#

Hence solutions are #p = (1+-3sqrt(5))/2#

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Answer 2

Julia Adams

#p=(1+3sqrt5)/2, (1-3sqrt5)/2#

#p^2−2p(1/2)+(1/4)-(11+1/4)=0 or, (p−1/2)^2=45/4=5(3/2)^2 or, [option 1] p−1/2 = 3sqrt5/2 or, p=(1+3sqrt5)/2 or, [option 2] p−1/2 = -3sqrt5/2 or, p=(1-3sqrt5)/2#

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Answer 3

Abigail Allen

To solve the quadratic equation 0 = p^2 - p - 11, you can use the quadratic formula, which is given by:

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

Where a, b, and c are the coefficients of the quadratic equation (ax^2 + bx + c = 0). In this equation, a = 1, b = -1, and c = -11.

Plugging these values into the quadratic formula:

[p = \frac{{-(-1) \pm \sqrt{{(-1)^2 - 4(1)(-11)}}}}{{2(1)}}]

[p = \frac{{1 \pm \sqrt{{1 + 44}}}}{2}]

[p = \frac{{1 \pm \sqrt{45}}}{2}]

So the solutions for p are:

[p = \frac{{1 + \sqrt{45}}}{2}]

[p = \frac{{1 - \sqrt{45}}}{2}]

These are the two solutions for the quadratic equation (0 = p^2 - p - 11).

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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.