When a common formula applies, it's always good to start by writing down the formula.moneyman wrote:If two of the four expressions x+y,x+5y,x-y and 5x-y are chosen at random , what is the probability that their product will be of the form x^2-(by)^2, where b is an integer??
1/2
1/3
1/4
1/5
1/6
I'm also having trouble with this problem. 4 choose 2 gives the number of unordered 2-groups, right? (this may be incorrect- please let me know if I'm misremembering) If so, the answer should be (c) because both {(a+b),(a-b)} and {(a-b),(a+b)} are viable subsets...Stuart Kovinsky wrote:When a common formula applies, it's always good to start by writing down the formula.moneyman wrote:If two of the four expressions x+y,x+5y,x-y and 5x-y are chosen at random , what is the probability that their product will be of the form x^2-(by)^2, where b is an integer??
1/2
1/3
1/4
1/5
1/6
Here, we know that:
probability = (# of desired outcomes)/(total # of possibilities)
We're choosing 2 items out of 4. So, the total # of possibilities is:
4C2 = 4!/2!2! = 24/4 = 6.
Therefore, the basic denominator is 6. Eliminate 1/4 and 1/5 (the answer could be 1/2 or 1/3 after cancelling).
Next, we need to calculate the # of products that can be written as the question demands.
If we recognize the new expression as a difference of squares (a^2 - b^2) our life becomes easier. The only way to form a difference of squares is to multiply:
(a + b) and (a - b).
The only two expressions in this form are (x+y) and (x-y). So, there's only one pair of expressions that give us what we want.
Therefore, 1 out of 6 possible outcomes matches the requirement: choose 1/6.
Stuart:Stuart Kovinsky wrote:
Next, we need to calculate the # of products that can be written as the question demands.
If we recognize the new expression as a difference of squares (a^2 - b^2) our life becomes easier. The only way to form a difference of squares is to multiply:
(a + b) and (a - b).
The only two expressions in this form are (x+y) and (x-y). So, there's only one pair of expressions that give us what we want.
Therefore, 1 out of 6 possible outcomes matches the requirement: choose 1/6.
The correct answer will be in the form of (x+y)(x-y); x and y can represent anything.ssuarezo wrote:Stuart:Stuart Kovinsky wrote:
Next, we need to calculate the # of products that can be written as the question demands.
If we recognize the new expression as a difference of squares (a^2 - b^2) our life becomes easier. The only way to form a difference of squares is to multiply:
(a + b) and (a - b).
The only two expressions in this form are (x+y) and (x-y). So, there's only one pair of expressions that give us what we want.
Therefore, 1 out of 6 possible outcomes matches the requirement: choose 1/6.
One question: Why did you assume coeficient=1 (a+b) and (a-b)? The stems tells x^2 - by^2, and that b is an integer, different from 1, as in the case of x, so ... why not to choose (x+5y) (x-5y)??? .. I know th answer is the same, but, that will not always be the case.
Thank you,
Silvia.
