SOLUTION: answer this please. a number is increased by two the square root of this result is less than the square root of half the number is 3. find the number.

Question 931147: answer this please. a number is increased by two the square root of this result is less than the square root of half the number is 3. find the number.
Answer by ptfile(81) (Show Source): You can put this solution on YOUR website!
Let x = The Number
A number is increased by two the square root of this result is less than the square root of half the number is 3.
√(x+2)-√(x/2)=3
√(x+2)=√(x/2)+3
(√(x+2))^2=((√(x/2)+3)^2
x+2=x/2+6√(x/2)+9
0 =-x-2+x/2+6√(x/2)+9
0 =-x/2+6√(x/2)+7
0 =-x/2+(6√(x/2))(√(2/2))+7
0 =-x/2+6√2x/2+7
0 =-x/2+6√2x/2+14/2
0 =-x+6√2x+14
x-14 = 6√2x
(x-14)^2= (6√2x)^2
x^2-28x+196= 72x
x^2-28x+196-72x=0
x^2-100x+196=0
Use Quadratic formula to solve for x

x=98
or
x=2
Substitute x for 98
√(98+2)-√(98/2)=3
√100-√49=3
10-7=3
3=3
Substitute x for 2
√(2+2)-√(2/2)=3
√(4)-√(1)=3
2-1=3
1=3
The number is 98.
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