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x and y are consecutive odd integers. If x^2 + y^2 = 394, what is the

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Re: x and y are consecutive odd integers. If x^2 + y^2 = 394, what is the [#permalink]

y =x +2

therefore, x^2 + (x+2)^2 = 394
which equates to
(x-13)(x+15) = 0

Either x= 13; then y= 15 ; xy = 195

Or x= -15; then y= -13 ; xy = 195
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Re: x and y are consecutive odd integers. If x^2 + y^2 = 394, what is the [#permalink]
Bunuel wrote:
x and y are consecutive odd integers. If x^2 + y^2 = 394, what is the value of product xy?

A. 143
B. 195
C. 255
D. 283
E. 321

x, y are consecutive odd integers => y=x+2

$$x^2+y^2=394$$
$$x^2+(x+2)^2=394$$
$$x^2+x^2+4x+4=394$$
$$2x(x+2)=394-4$$
$$x(x+2)=\frac{390}{2}$$
$$x(x+2)=195$$
$$xy=195$$

B.
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Re: x and y are consecutive odd integers. If x^2 + y^2 = 394, what is the [#permalink]
Bunuel wrote:
x and y are consecutive odd integers. If x^2 + y^2 = 394, what is the value of product xy?

A. 143
B. 195
C. 255
D. 283
E. 321

x = 2k-1
y = 2k+1

=> (2k-1)^2 + (2k+1)^2 = 394
solving, k^2 = 49

so, xy = (2k+1)(2k-1) = (2k)^2 - (1)^2 = 195. B
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Re: x and y are consecutive odd integers. If x^2 + y^2 = 394, what is the [#permalink]
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Bunuel wrote:
x and y are consecutive odd integers. If x^2 + y^2 = 394, what is the value of product xy?

A. 143
B. 195
C. 255
D. 283
E. 321

Since x and y are consecutive odd integers, we can say:

x = y + 2

x - y = 2

Squaring both sides of our equation, we have:

(x - y)^2 = 2^2

x^2 + y^2 - 2xy = 4

x^2 + y^2 = 4 + 2xy

Since x^2 + y^2 = 4 + 2xy, we can substitute 4 + 2xy for x^2 + y^2 in the equation x^2 + y^2 = 394 and we have:

4 + 2xy = 394

2xy = 390

xy = 195 (which is the product of x and y)