$$If\ x+y=a\ and\ x-y=b,\ then\ 2xy=?$$
$$\left(a\right)\ \frac{\left(a^2-b^2\right)}{2}$$
$$\left(b\right)\ \frac{\left(b^2-a^2\right)}{2}$$
$$\left(c\right)\ \frac{\left(a^{ }-b\right)}{2}$$
$$\left(d\right)\ \frac{\left(ab\right)}{2}$$
$$\left(e\right)\ \frac{\left(a^2+b^2\right)}{2}$$
OA is a
How do i set up the formula here to get the right answer to this question?
Thank you for your help
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Hi Roland2rule,
This question can be approached in a couple of different ways, Here's how you can TEST VALUES to get to the answer:
IF...
X=3 and Y=2
A=5 and B=1
So we're looking for an answer that equals (2)(3)(2) = 12 when A=5 and B=1. There's only one answer that matches...
Final Answer: A
GMAT assassins aren't born, they're made,
Rich
This question can be approached in a couple of different ways, Here's how you can TEST VALUES to get to the answer:
IF...
X=3 and Y=2
A=5 and B=1
So we're looking for an answer that equals (2)(3)(2) = 12 when A=5 and B=1. There's only one answer that matches...
Final Answer: A
GMAT assassins aren't born, they're made,
Rich
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Hi Roland2rule,$$If\ x+y=a\ and\ x-y=b,\ then\ 2xy=?$$
$$\left(a\right)\ \frac{\left(a^2-b^2\right)}{2}$$
$$\left(b\right)\ \frac{\left(b^2-a^2\right)}{2}$$
$$\left(c\right)\ \frac{\left(a^{ }-b\right)}{2}$$
$$\left(d\right)\ \frac{\left(ab\right)}{2}$$
$$\left(e\right)\ \frac{\left(a^2+b^2\right)}{2}$$
OA is a
How do i set up the formula here to get the right answer to this question?
Thank you for your help
Lets take a look at your question,
We can use this problem using the formula:
$$\left(x+y\right)^2-\left(x-y\right)^2=4xy$$
Plugin the given values,
$$a^2-b^2=4xy$$
$$a^2-b^2=2\times2xy$$
$$\frac{a^2-b^2}{2}=2xy$$
$$2xy=\frac{a^2-b^2}{2}$$
Therefore, option A is correct.
Hope it helps.
I am available if you'd like any follow up.
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Squaring both equations, we have:BTGmoderatorRO wrote:$$If\ x+y=a\ and\ x-y=b,\ then\ 2xy=?$$
$$\left(a\right)\ \frac{\left(a^2-b^2\right)}{2}$$
$$\left(b\right)\ \frac{\left(b^2-a^2\right)}{2}$$
$$\left(c\right)\ \frac{\left(a^{ }-b\right)}{2}$$
$$\left(d\right)\ \frac{\left(ab\right)}{2}$$
$$\left(e\right)\ \frac{\left(a^2+b^2\right)}{2}$$
OA is a
How do i set up the formula here to get the right answer to this question?
Thank you for your help
(x + y)^2 = a^2
x^2 + y^2 + 2xy = a^2
and
(x - y)^2 = b^2
x^2 + y^2 - 2xy = b^2
Subtracting the second equation from the first, we have:
4xy = a^2 - b^2
2xy = (a^2 - b^2)/2
Answer: A
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(x+y)^2 = x^2+y^2+2xyBTGmoderatorRO wrote:$$If\ x+y=a\ and\ x-y=b,\ then\ 2xy=?$$
$$\left(a\right)\ \frac{\left(a^2-b^2\right)}{2}$$
$$\left(b\right)\ \frac{\left(b^2-a^2\right)}{2}$$
$$\left(c\right)\ \frac{\left(a^{ }-b\right)}{2}$$
$$\left(d\right)\ \frac{\left(ab\right)}{2}$$
$$\left(e\right)\ \frac{\left(a^2+b^2\right)}{2}$$
OA is a
How do i set up the formula here to get the right answer to this question?
Thank you for your help
(x-y)^2 = x^2+y^2-2xy
a^2=x^2+y^2+2xy is equation 1
b^2 =x^2+Y^2-2xy is equation 2
Subtracting equation 2 from equation 1 will give
a^2-b^2 = 4xy
2xy = (a^2-b^2)/2
