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

|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

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.

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

Hi Albs

the highlightedvalue 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.

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

\(? = xy\)


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


The correct answer is therefore (D).


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

Regards,
Fabio.

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

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

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

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

Therefore, the answer is D.
Answer: D

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

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.

x*y = 7*5 = 35.

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

Since (x-y)^2>=0
- |y-5|>=0
|y-5|<=0
|y-5|=0
y = 5
x = 7
xy = 5*7 = 35

IMO D

Kindly see the attachment.

I personally don't think this question is a 700+ level question. After the concept that square and modulus will both be +ve, there is nothing left to the question. If there was further calculations or some manipulations required, then it could have been a 700+ Level question.
MathRevolution, kindly see if this question needs to be re-evaluated.

Like many point out here.

(x-7)^2 is always greater than or equal to 0
-|y-5| is always less than or equal to 0

Therefore they can only equal each other when they are both zero.

(x-7)^2=0 when x=7
-|y-5|=0 when y=5

Therefore xy=7*5=35

Answer: D

(x−7)^2=−|y−5| this equality is valid only if
-|y-5|=0 => y=5
and (x-7)^2 = 0 => x=7
=> x*y=35
the correct answer is D

I dont think calculation is required at all. Square is always positive. From the given equation \((x-7)^2\) = -|y-5|, only value that satisfies it is 0. Hence, x=7 and y=5

I wonder, how the square of something can be negative.