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![How many integers satisfy the following equation ? [square_root][[squa](/forum/styles/gmatclub_light/imageset/icon_post_target_unread.svg "How many integers satisfy the following equation ? [square_root][[squa"){kind=icon}
03 Apr 2020, 23:46
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Difficulty:
95%
(hard)
Question Stats:
15% (02:28) correct
85%
(02:25)
wrong
based on 261
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How many integers satisfy the following equation ?
\(\sqrt{[(x-1)^{1/2} -2]^2} + \sqrt{[(x-1)^{1/2} -3]^2} = 1\)
A. 0
B. 1
C. 2
D. 4
E. 6
\(\sqrt{[(x-1)^{1/2} -2]^2} + \sqrt{[(x-1)^{1/2} -3]^2} = 1\)
A. 0
B. 1
C. 2
D. 4
E. 6
04 Apr 2020, 19:04
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Always keep 'plugging values' as your last option or if you're short of time.
let \((x-1)^{1/2} = a\)
\(\sqrt{[a -2]^2} + \sqrt{[a -3]^2} = 1\)
|a-2| + |a-3| = 1
Case 1- a<2
|a-2| + |a-3| = 1
2-a+3-a =1
a=2 (rejected, as it doesn't lie in the range)
Case 2- 2≤a≤3
|a-2| + |a-3| = 1
a-2+3-a =1
1=1
each real number in the range satisfies the equation
Case 3- a>3
|a-2| + |a-3| = 1
a-2+a-3 = 1
a=3 (rejected)
Solution of |a-2| + |a-3| = 1 is 2≤a≤3
\(2≤(x-1)^{1/2}≤3\)
4≤(x-1)≤9
5 ≤ x ≤ 10
integers that satisfy the following equation are 5, 6, 7, 8, 9 and 10
let \((x-1)^{1/2} = a\)
\(\sqrt{[a -2]^2} + \sqrt{[a -3]^2} = 1\)
|a-2| + |a-3| = 1
Case 1- a<2
|a-2| + |a-3| = 1
2-a+3-a =1
a=2 (rejected, as it doesn't lie in the range)
Case 2- 2≤a≤3
|a-2| + |a-3| = 1
a-2+3-a =1
1=1
each real number in the range satisfies the equation
Case 3- a>3
|a-2| + |a-3| = 1
a-2+a-3 = 1
a=3 (rejected)
Solution of |a-2| + |a-3| = 1 is 2≤a≤3
\(2≤(x-1)^{1/2}≤3\)
4≤(x-1)≤9
5 ≤ x ≤ 10
integers that satisfy the following equation are 5, 6, 7, 8, 9 and 10
preetamsaha
04 Apr 2020, 22:45
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nick1816
Very well explained by nick1816. I'll explain through the number line which should make your understanding a bit more easier.
\(\sqrt{[(x-1)^{1/2} -2]^2} + \sqrt{[(x-1)^{1/2} -3]^2} = 1\)
Let \((x-1)^{\frac{1}{2}}=y\).
Then, \(\sqrt{[y -2]^2} + \sqrt{[y -3]^2} = 1.....|y-2|+|y-3|=1\)
Now look at the attached figure..
Case (I) :- When y<2..
In this case the answer will be \(|y-2|+|y-3|=(2-y)+(3-y)=2-y+3-y=5-2y\). As y<2, 5-2y>1 always
Case (II) :- When y is between 2 and 3 including the extremes..
\(2\leq{y}\leq{3}\)
In this case the answer will always be 1....since \(|y-2|+|y-3|=(y-2)+(3-y)=y-2+3-y=1\)
Case (III) :- When y>3..
In this case the answer will be \(|y-2|+|y-3|=(y-2)+(y-3)=y-2+y-3=2y-5\). As y>3, 2y-5>1 always
So ONLY case (II)..
\(2\leq{y}\leq{3}\).....\(2\leq{\sqrt{x-1}}\leq{3}............4\leq{x-1}\leq{9}......4+1\leq{x}\leq{9+1}\)...
\(5\leq{x}\leq{10}\)
As x is an integer, x= 5, 6, 7, 8, 9 and 10.
Thus 6 values...
E
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