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What is the largest possible value of y if (x+3)^2 + (y+6)^2 = 100?

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What is the largest possible value of y if (x+3)^2 + (y+6)^2 = 100?  [#permalink]

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22 Aug 2018, 00:49
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25% (medium)

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76% (01:18) correct 24% (02:25) wrong based on 33 sessions

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What is the largest possible value of y if (x+3)^2 + (y+6)^2 = 100?

A. -2
B. 4
C. 5
D. 8
E. 12

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Re: What is the largest possible value of y if (x+3)^2 + (y+6)^2 = 100?  [#permalink]

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22 Aug 2018, 01:00
Let's consider X = -3. Then the max value of Y can be 4.

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What is the largest possible value of y if (x+3)^2 + (y+6)^2 = 100?  [#permalink]

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22 Aug 2018, 01:02

Solution

Given:
• Two variables x and y
• $$(x+3)^2 + (y+6)^2 = 100$$

To find:
• Largest possible value of y

Approach and Working:
If y is largest, then $$(y+6)^2$$ must be largest, and simultaneously $$(x+3)^2$$ must be smallest.
• Now, the minimum value of any square number is always 0.
o Hence, if y is largest, we can say $$(x+3)^2$$ is 0.
o Therefore, $$(y+6)^2 = 100 = 10^2$$
Or, y + 6 = 10 (ignoring the negative root, as we are maximising y)
Or, y = 4

Hence, the correct answer is option B.

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Re: What is the largest possible value of y if (x+3)^2 + (y+6)^2 = 100?  [#permalink]

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22 Aug 2018, 01:20
Bunuel wrote:
What is the largest possible value of y if (x+3)^2 + (y+6)^2 = 100?

A. -2
B. 4
C. 5
D. 8
E. 12

Given

(x+3)^2 + (y+6)^2 = 100

we are looking for the largest value of y. Therefore we need to minimize $$(x + 3)^2$$. x can be anything as there are no conditions related to x are given. any value other than - 3 is not accepted in this case.

$$(- 3 + 3 )^2 + (y + 6)^2 = 100$$

$$(0)^2 + ( 4 + 6 )^2 = 100$$
So, y = 4.

anyone can plug other values but these values won't work as we have even exponents .

Re: What is the largest possible value of y if (x+3)^2 + (y+6)^2 = 100? &nbs [#permalink] 22 Aug 2018, 01:20
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