How many integers satisfy the following equation ? [square_root][[squa

Question

How many integers satisfy the following equation ?

\sqrt{ ^2} + \sqrt{ ^2} = 1

A. 0
B. 1
C. 2
D. 4
E. 6

nick1816 · Accepted Answer

Always keep 'plugging values' as your last option or if you're short of time.

let  (x-1)^{1/2} = a

\sqrt{ ^2} + \sqrt{ ^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

chetan2u · Answer

Very well explained by nick1816 . I'll explain through the number line which should make your understanding a bit more easier. \sqrt{ ^2} + \sqrt{ ^2} = 1 Let (x-1)^{\frac{1}{2}}=y . Then, \sqrt{ ^2} + \sqrt{ ^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

preetamsaha · Answer

here,( x-1 ) must be ≥ 0 for (x-1)^1/2 to be validated .
plugging values of x, only satisfies at x=5 and 10 ,
so no of integral solution of x = 2 
 correct answer is c

ArunSharma12 · Answer

as both the expressions are under square root, they will be either 0 or more than 0.
we want one of the expressions to give 0 and other expression to give 1.
consider x = 5, 
 \sqrt{ ^2} = \sqrt{ ^2} = 0 
 \sqrt{ ^2} = \sqrt{ ^2} = 1

consider x = 10,
 \sqrt{ ^2} = \sqrt{ ^2} = 1 
 \sqrt{ ^2} = \sqrt{ ^2} = 0

thus, two integers 5 and 10 satisfies the expression.
Ans: C

KarishmaB · Answer

Recall that  \sqrt{x^2} = |x|

So  \sqrt{ ^2} + \sqrt{ ^2} = 1 
becomes  |\sqrt{x-1} - 2| + |\sqrt{x-1}-3| = 1

Put  \sqrt{x-1} = y

So |y - 2| + |y - 3| = 1
Distance of y from 2 + Distance of y from 3 = 1
Since distance between 2 and 3 is 1, y could lie anywhere between 2 and 3 inclusive.

2 \leq y \leq 3

\sqrt{x-1} = 2 
x = 5

\sqrt{x-1} = 3 
x = 10

So for all values of x between 5 and 10 inclusive, y will lie between 2 and 3 inclusive. 
Hence x has 6 integer values.

Answer (E)

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How many integers satisfy the following equation ? [square_root][[squa

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