If x and y are numbers such that (x + 11) (y – 11) = 0, what is the sm

Should be D.
Either x or y has to be |11|. For the expression to be smallest, we consider the other as 0. Hence, 0^2 + 11^2 = 121.

Given that,

(x+11)(y-11) =0

so either x = -11 or y=11 or both

But we have to minimise X^2 + Y^2.

That can be minimised only when one of x or y is zero

so if we take x=-11

then X^2 + Y^2
=(-11)^2 + 0
=121

So answer should be D

Take x=-11,y=0 OR y=11, x=0
Smallest possible value= -11*2+0^2
OR O^2+11^2=121
Answer D

VERITAS PREP OFFICIAL SOLUTION

The first thing to recognize is that one of these answers is the correct answer. This means that, given the parameters of the problem, one of these choices should spit out an answer that makes sense if used in the above equations. The easiest thing to do to check which answer choice will give a plausible solution is to plug one of the choices in to our given equations.

A) 0

x² + y² = 0 (Plausible, both x and y must be 0)

(0 + 11)(0 – 11) = 0 (not possible)

By plugging in our first answer choice we are given some very important information. We see that in the equation “(x + 11) (y – 11) = 0”, one or both of the quantities within the parentheses must be zero. That is one of those rules of algebra that we all learned a million years ago in our first algebra class called the zero- product property : If ab = 0, then a = 0 or b = 0. Given this fact, we need x to equal -11 or y to equal 11 in order for the first equation to be possible. This will help us in testing our next answer choices.

B) 11

x² + y² = 11 (Not plausible given we need x to equal -11 or y to equal 11)

C) 22

x² + y² = 22 (Not plausible given we need x to equal -11 or y to equal 11)

D) 121

x² + y² = 121 (Plausible, x² could equal 0 and y could equal 11, or x could equal -11 and y could equal 0)

(-11 + 11)(0 – 11) = 0 or (0 + 11)(11 – 11) = 0 (Plausible)

At this point we are actually done. The answer choices are always listed in order of smallest to largest and we are looking for the smallest number that would satisfy these parameters. Even if (E) gives us a plausible answer, it is a larger number than (D) and, thus, is an incorrect answer.

There are a number of different kinds of problems where plugging in the answer choices are useful, but these tend to be problems where either an equation or some parameters are given and the goal is to find out which answer choice fits given what is stated in the problem. Even if plugging in an answer choice doesn’t give an immediate answer, it can shed some light onto some aspect of the problem that might not be immediately visible, as we saw above. So go ahead and use those answer choices. They are there to help you – not to hurt you!

From (x + 11)(y - 11)=0 it follows that either x = -11 or y = 11. Thus either x^2 = 121 or y^2 = 121.

Now, if x^2 = 121, then the least value of y^2 is 0, so the least value of x^2 + y^2 = 121 + 0 = 121.
Similarly if y^2 = 121, then the least value of x^2 is 0, so the least value of x^2 + y^2 = 0 + 121 = 121.

Answer: D.

Similar question to practice: if-x-and-y-are-numbers-such-that-x-9-y-9-0-what-is-the-159800.html

Love your way of thinking. Could you please confirm my understanding of these kind of questions - if (x+11)(y-11) = 0, then any of the following 3 scenarios are possible -

1. x= -11 and y can be anything
2. y = 11 and x can be anything
3. both x is -11 and y is 11

Please confirm. Thanks a ton.

Since, \((x + 11) (y – 11) = 0\)

Either -

1. \((x + 11) = 0\) or, \(x = - 11\) & \(y = 0\)
2. \((y – 11) = 0 , or y = 11\) & \(x = 0\)

Now, \(x^2 + y^2\) can have 2 values -

1. \(-11^2 + 0^2 = 121\)
2. \(0^2 + 11^2 = 121\)

Thus, the smallest possible value of \(x^2 + y^2\) will be (D) \(121\)

Easy although I failed to get it
what is the least value which can make below mentioned equation Zero
(x+11)(y-11)=0
either put x=-11 or y=11 it will make the other zero and put zero for other variable
so x=-11,y=0
result is 0
now find value of x^2+y^2
(-11)^2+(0)^2
121
choice is D

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