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Updated on: 23 Jul 2025, 13:53
Originally posted by Alhamdan1995 on 16 May 2020, 14:59.
Last edited by Krunaal on 23 Jul 2025, 13:53, edited 1 time in total.
Last edited by Krunaal on 23 Jul 2025, 13:53, edited 1 time in total.
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
71% (01:51) correct
29%
(01:50)
wrong
based on 559
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Which of the following is a set of all possible values of y that satisfy \(y-13\sqrt{y}+36 = 0\)
A) (2,3)
B) (+/-2,+/-3)
C) (4,9)
D) (-4,-9)
E) (16,81)
A) (2,3)
B) (+/-2,+/-3)
C) (4,9)
D) (-4,-9)
E) (16,81)
16 May 2020, 20:52
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Alhamdan1995
Choices
Since we have \(\sqrt{y}\), y cannot be negative, so you can straight way eliminate the choices B and D.
A) (2,3)
\(y-13\sqrt{y}+36 = 0\) => \(2-13\sqrt{2}+36 = 0\)..NO
B) (4,6)
\(y-13\sqrt{y}+36 = 0\) => \(4-13\sqrt{4}+36 = 0.....40-26=0\)..NO
C) (16,81)
\(y-13\sqrt{y}+36 = 0\) => \(16-13\sqrt{16}+36 = 0....42-13*4=0\)..YES
Algebraic
Let \(\sqrt{y}=x..\)
So, \(x^2-13x+36=0...(x-4)(x-9)=36\)
But, remember we are looking for y, and \(y=x^2...so y=4^2=16....y=9^2=36\)
E
General Discussion
17 May 2020, 08:13
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Alhamdan1995
\(y-13\sqrt{y}+36 = 0\)
Let \(\sqrt{y}\) be x; y=x^2
\(x^2 - 13x + 36 = 0\)
\(x^2 - 9x - 4x + 36 = 0\)
(x-9)(x-4)=0
x = 9 or x = 4
If x=9; \(\sqrt{y} = 9; y=81\)
If x=4; \(\sqrt{y} = 4; y=16\)
IMO E
