If (x - 7)^2 = -|y - 5|, xy = ?

Question

If  (x-7)^2=-|y-5|  ,  xy= ?
 
A. 5

B. 7

C. 12

D. 35

E. cannot be determined

chetan2u · Accepted Answer

A square and a modulus can never be NEGATIVE 
So least value of a square or modulus is 0..

(x-7)^2=-|y-5|  ..
Here -|y-5| will never be positive as |y-5| is greater than or equal to 0..
A square on left side tells us that (x-7)^2=0, or x=7..
And -|y-5|=0 means y =5
Therefore xy=7*5=35

D

Chethan92 · Answer

If y>5, then -|y-5| = -(y-5) = 5-y
 x^2+49-14x = 5-y 
 x^2+44-14x+y = 0 ........(1)
If y<5, then -|y-5| = -(-y+5) = y-5
 x^2+49-14x = y-5 
 x^2+54-14x = y ..........(2)
(1) in (2) gives
 2x^2+98-28x = 0 
 x^2+49-14x = 0 
 (x-7)^2 = 0 
x = 7
If x = 7, then -|y-5| = 0; y = 5
xy = 35

D is the answer.

Archit3110 · Answer

|y-5|= sqrt ( y-5)^2

and 
(x-7)^2 = sqrt ( (x-5)^2
squaring both sides
(x-7) = (x-5)^2
solve
y^2-10y+32-x=0-- ( 1)

seeing answer option
x*y=35
x=7 & y = 5 
satisfies the eqn ( 1) 
so IMO C

Albs · Answer

I assume

x=7 because (x-7)=0, therefore x= 7 when you rearrange

then
7= -|y-5|
7= -(y-5)
7= -y+5
y= -2

Therefore x*y= -14

Is it E based on my assumption above? I don't think so but made sense to me.

KanishkM · Answer

We can start with plugging in the values for x and y

Only D when plugged in the  (x-7)^2=-|y-5| , gives equal values

1 * 5 or 5 * 1
1 * 7 or 7 * 1
3 * 4 or 4 * 3 or 2 * 6 or 6 * 2

KanishkM · Answer

Hi  Albs

the  highlighted value when substituted back in the  bold   part should had satisfied the modulus, which it doesn't do, Therefore the value cannot be applicable to this case.

so you got y = -2

lets put it back in  7= -|y-5|

7 ! = - 7

You have to search for different cases for x and y to satisfy (x−7)^2=−|y−5|

But then you are provided with the product of x and y, just substitute them in place of x and y and you should be good.

fskilnik · Answer

{\left( {x - 7} \right)^2} = - \left| {y - 5} \right|\,\,\,\,\,\left( * \right)

? = xy

\left. \matrix{
 {\left( {x - 7} \right)^2} \ge 0\,\,\, \hfill \cr 
 - \left| {y - 5} \right| \le 0 \hfill \cr} \right\}\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,{\left( {x - 7} \right)^2} = 0 = - \left| {y - 5} \right|\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left\{ \matrix{
 \,x - 7 = 0 \hfill \cr 
 \,y - 5 = 0 \hfill \cr} \right.\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 5 \cdot 7 = 35

The correct answer is therefore (D).

We follow the notations and rationale taught in the  GMATH  method.

Regards, 
Fabio.

rish2708 · Answer

I have a query.

|y-5| will always be +ve an (x-7)^2 = -|y-5| can't be possible as LHS is a square number?
So how can we find a deterministic answer in such case?

Regards,
Rishav

rish2708 · Answer

Got it.. only 0 it can satisfy!! Thanks

kunalcvrce · Answer

Since Square of any number is >=0;  A^2>=0 
 (x-7)^2=A,-|y-5|=B 
since A cant be negative ,So B=0
y=5 and x=7
X*Y=35

MathRevolution · Answer

=>
 (x-7)^2=-|y-5| 
 => (x-7)^2+|y-5| = 0 
 => x = 7  and  y = 5 
This yields  xy = 35 .

Therefore, the answer is D.
Answer: D

Vinit800HBS · Answer

MathRevolution

Sorry to say but your answer gives no additional knowledge and thus is not really useful in this case.

The point to observe is that a perfect square has been equated a negative absolute value function.

We must note that a perfect square can never be negative. It can only achieve a value of zero or positive. Same is the case with absolute value function.

This, to ensure that both the functions yield the same answer, we must equate both the function to zero. 
This, x=7 and y=5

XY = 35

Posted from my mobile device

MathRevolution · Answer

If x = 7, then we have (x-7)^2 = -|y-5| or (7-7)^2 = -|y-5|.
Thus -|y-5| = 0 and we have y = 5.