If p^2 – 13p + 40 = q, and p is a positive integer between 1

Thnkx a ton bunuel for quick solution.Excellent!!
Cheers:P

(p-5)(p-8) = q

p = 6 or 7 for this to be true, so

2/10 = 1/5

P is any integer between 1-10, inclusive.
Plug in the each values. It took me about 1:30 min to substitute all the values.

I did this because I can't visualize the graph of the equation other than I know it will be a parabola because of the X^2.

When p = 1, 2, 3, 4, 5, 8, 9, 10 the value of q is not negative.
Only time q is negative is when p = 6 or 7.
Probability 2/10 = 1/5 = B

Is this question not just easy enough and factor and then plug in numbers??

if: \(p^2-13p+40\) Factors out to (p-8)(p-5)=q, which can easily be done in your head very quickly.

Then make a list of numbers that would give a negative result.

It can only be 2 numbers 6 and 7, since anything over 7 yields either 0 or a positive number, and anything less than 6 yields a positive number, there can only be two cases which would satisfy the solution. Is making a graph really necessary?
Making it 2/10 or 1/5

P(P-13) plus 40 = q

while testing the values q is negative only when p is 7 or 6

thus the answer is 2/10 or 1/5

Given: \(p^2 – 13p + 40\) = q, 1< p < 10
Required: Probability of q < 0

\(p^2 – 13p + 40\) = (p -8)(p-5) = q

We need q < 0
Hence
(p -5)(p-8) < 0

Hence 5< p < 8.
Only two integral values lie in the range: 6 and 7

Probability = 2/10 = 1/5
Option B.

Solving an inequality with a less than sign:
The value of the variable will be greater than the smaller value and smaller than the greater value.
i.e. It will lie between the extremes.

Solving an inequality with a greater than sign:
The value of the variable will be smaller than the smaller value and greater than the greater value.
i.e. It can take all the values except the values in the range.

The quadratic equation could be expanded to (p-5)(p-8)=q
If p is a member of the set [1,10], p could be one of the 10 choices.

The equation (p-5)(p-8) will work out to < 0 for the range (5,8). The integers between 5 and 8 exclusive are 6 and 7.

The probability that q <0 = probability(p = 5 or p = 8 is selected) = 2/10 = 1/5

Factor the quadratic:
p^2 – 13p + 40 = q
(p – 8)(p – 5) = q
For p = 5 and p = 8, q = 0. Between p = 5 and p = 8, q has a negative sign, as (p – 8) is negative and (p – 5) is positive. With a total of 10 possible integer p values, only two (p = 6 and p = 7) fall in the range 5 < p < 8, so the probability is 2/10 or 1/5.
The correct answer is B.

We can factor the given equation:

p^2 – 13p + 40 = q

(p - 5)(p - 8) = q

We see that in order for q to be negative, either (p - 5) is negative and (p - 8) is positive OR (p - 5) is positive and (p - 8) is negative.

Analyzing our expression a bit further, we see that it only produces a negative product when p = 6 and p = 7.

Thus, the probability that q < 0 is 2/10 = 1/5.

Answer: B

Thanks for showing the explanation using the parabola Bunuel.

Was curious to understand how will it look like if for eg the question put the condition of q>0. In that case, the quadratic would turn out to be (p-8)(p-5) > 0. In such a case the parabola will open downward but then how do we make this out as the constant a (in a*p^2) is in any case +ve..

p² - 13p + 40 = 0
(p-5)(p-8) = 0

The CRITICAL POINTS are p=5 and p=8.
Only at these two critical points does p² - 13p + 40 = 0.
To determine the ranges in which p² - 13p + 40 < 0, test one value to the left and right of each critical point.

There are three ranges to consider:
p < 5, 5 < p < 8, p > 8
If p=0 or p=100, then p² - 13p + 40 is clearly GREATER than 0, implying that p<5 and p>8 are not viable ranges.
Implication:
The only viable range is 5 < p < 8.
Thus, of the 10 integers between 1 and 10, inclusive, only two -- 6 and 7 -- will yield a negative for p² - 13p + 40:
\(\frac{2}{10} = \frac{1}{5}\)

If we factor the right side of the equation, we can come up with a more meaningful relationship between p and q: p2 – 13p + 40 = q so (p – 8)(p – 5) = q.
We know that p is an integer between 1 and 10, inclusive, so there are ten possible values for p. We see from the factored equation that the sign of q will depend on the value of p.
One way to solve this problem would be to check each possible value of p to see whether it yields a positive or negative q.
However, we can also use some logic here.
For q to be negative, the expressions (p – 8) and (p – 5) must have opposite signs.
Which integers on the number line will yield opposite signs for the expressions (p – 8) and (p – 5)?
Those integers in the range 5 < p < 8 (notice 5 and 8 are not included because they would both yield a value of zero and zero is a nonnegative integer).
That means that there are only two integer values for p, 6 and 7, which would yield a negative q. With a total of 10 possible p values, only 2 yield a negative q, so the probability is 2/10 or 1/5.
The correct answer is B.

Factor the quadratic:

p^2 - 13p + 40 = q
(p - 5) (p - 8) = q

What is the probability that q < 0?

Since we know that p is a positive integer between 1 and 10, inclusive, then we can plug in numbers to check.

If q is negative, we know either (p - 5) or (p - 8) will be negative. We can quickly determine that if p is 6 or 7, q will be negative.

2 / 10 = 1/5

Answer is B.

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