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805+ (Hard)|   Algebra|      
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Bunuel
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If y + 4 = (x - 2)^2, x + 4 = (y - 2)^2, and x ≠ y, what is the value 25 Apr 2019, 02:20
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33% (03:02) correct 67% (03:02) wrong based on 130 sessions

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If \(y + 4 = (x - 2)^2\), \(x + 4 = (y - 2)^2\), and \(x ≠ y\), what is the value of \(x^2 + y^2\)?

(A) 10
(B) 15
(C) 20
(D) 25
(E) 30
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B
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jfranciscocuencag
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If y + 4 = (x - 2)^2, x + 4 = (y - 2)^2, and x ≠ y, what is the value 25 Apr 2019, 19:07
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Could someone plz provide answer?
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yvankd
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If y + 4 = (x - 2)^2, x + 4 = (y - 2)^2, and x ≠ y, what is the value 25 Apr 2019, 19:39
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Bunuel
If \(y + 4 = (x - 2)^2\), \(x + 4 = (y - 2)^2\), and \(x ≠ y\), what is the value of \(x^2 + y^2\)?

(A) 10
(B) 15
(C) 20
(D) 25
(E) 30
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From the first formula, we get x^2 = y + 4x. From the second formula we get y^2 = x + 4y.
So x^2 + y^2 = y + 4x + (x + 4y)
If we simplify, it would be 5y + 5x. Since y ≠ x, we can eliminate A, C, and E, but I'm not sure how we can eliminate B or D. What's the OE?

Thank you.
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TornDilemma
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If y + 4 = (x - 2)^2, x + 4 = (y - 2)^2, and x ≠ y, what is the value 25 Apr 2019, 21:39
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Bunuel
If \(y + 4 = (x - 2)^2\), \(x + 4 = (y - 2)^2\), and \(x ≠ y\), what is the value of \(x^2 + y^2\)?

(A) 10
(B) 15
(C) 20
(D) 25
(E) 30
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\(y+4=(x-2)^2\) & \(x+4=(y-2)^2\)

Subtracting 2 from 1,

\(y-x=(x-2)^2-(y-2)^2\)
or \(y-x=(x-y)*(x+y-4)\)

or \(x+y-4=-1\)or\(x+y=3\)

Now, Simplifying 1 and 2,
\(x^2-4x=y\)& \(y^2-4y=x\)

Adding these, \(x^2+y^2=5(x+y)=5*3=15\).

IMo, Option B.
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If y + 4 = (x - 2)^2, x + 4 = (y - 2)^2, and x ≠ y, what is the value 26 Apr 2019, 00:24
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Subtract both equations and we will get
x^2-y^2-3x+3y= 0
(x+y)(x-y) -3 (x-y) =0
(x-y) (x+y-3) = 0, as x is not equal to y
x+y=3
Now add both equations, we will get
x^2+y^2= 5(x+y)=5*3=15
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If y + 4 = (x - 2)^2, x + 4 = (y - 2)^2, and x ≠ y, what is the value 26 Apr 2019, 12:26
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nick1816
Subtract both equations and we will get
x^2-y^2-3x+3y= 0
(x+y)(x-y) -3 (x-y) =0
(x-y) (x+y-3) = 0, as x is not equal to y
x+y=3
Now add both equations, we will get
x^2+y^2= 5(x+y)=5*3=15
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Hello nick1816!

How did you arrive to the red part?
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nick1816
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If y + 4 = (x - 2)^2, x + 4 = (y - 2)^2, and x ≠ y, what is the value 26 Apr 2019, 12:51
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y+4=\((x - 2)^2\)
y+4=\(x^2\)-4x+4
\(x^2\)-4x-y=0.......(1)
similarly solve x + 4 = \((y - 2)^2\), you will get
\(y^2\)-4y-x=0......(2)
Now subtract equation (2) from equation (1), you will get
\(x^2-y^2-3x+3y\)= 0



jfranciscocuencag
nick1816
Subtract both equations and we will get
x^2-y^2-3x+3y= 0
(x+y)(x-y) -3 (x-y) =0
(x-y) (x+y-3) = 0, as x is not equal to y
x+y=3
Now add both equations, we will get
x^2+y^2= 5(x+y)=5*3=15

Hello nick1816!

How did you arrive to the red part?
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If y + 4 = (x - 2)^2, x + 4 = (y - 2)^2, and x ≠ y, what is the value 28 Apr 2019, 11:58
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Eqn1: y+4=(x−2)^2
Eqn2: x+4=(y−2)^2

Step1: Subtract Eqn 1 from 2,
we get x-y= 3

Step2: Adding Eqn 1 and 2,
we get x^2+y^2= 5(x+y)

Substituting Step 1 in Step 2, we get x^2+y^2= 5*3= 15

Ans B

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If y + 4 = (x - 2)^2, x + 4 = (y - 2)^2, and x ≠ y, what is the value 01 May 2019, 11:26
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If y+4=(x−2)2y+4=(x−2)2, , and x≠yx≠y, what is the value of x2+y2
x+4=(y−2)2x+4=(y−2)2
Subtracted both the equations :
x+y = 3

Added both the equations
y+x +8 = (x-2)^2 - (y-2)^2
3+8 = x^2+y^2 +8 -4(x+y)
3+8 = x^2+y^2 +8 -4(3)
3 +12 =x^2 +y^2
x^2 +y^2 = 15

Option B
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If y + 4 = (x - 2)^2, x + 4 = (y - 2)^2, and x ≠ y, what is the value 06 Oct 2024, 06:20
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