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lvl10720-Turquoise
20-Turquoise
February 20, 2011
Solved

Irrational Number and Solving Equation.

  • February 20, 2011
  • 1 reply
  • 7017 views

Hello everyone,

I have a question and wish learning. It's about "Irrational Number and Solving Equation".

(1).PNG

(2).PNG

Thanks in advance.

Regards,

Loi,

Best answer by RichardJ

sqrt(x)-floor(sqrt(x)) is the non-integer part of sqrt(x). You know that is equal to the right hand side, or to sqrt(7)-2. So you can generate other solutions just be adding an integer to that number.

1 reply

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RichardJ19-Tanzanite
19-Tanzanite
February 20, 2011

Your equation requires that the decimal part of sqrt(x) be equal to the expression on the right hand side. However, since there is nothing to specify what the integer part of sqrt(x) has to be, your equation has an infinite number of possible solutions.

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lvl10720-TurquoiseAuthor
20-Turquoise
February 21, 2011

Thanks for your suggestion, Richard.

Following up the suggestion, above. Checking with x=6; x=7; x=8. We have:

substitute,x=6,False.PNGx=6,False.PNG

substitute,x=7,True.PNGx=7,True.PNG

subtitute,x=8,False.PNGx=8,False.PNG

So x=7 is the one of the solutions. And the rests of the solutions are the question.

Thanks in advance.

Regards,

Loi.

R
RichardJ19-TanzaniteAnswer
19-Tanzanite
February 21, 2011

sqrt(x)-floor(sqrt(x)) is the non-integer part of sqrt(x). You know that is equal to the right hand side, or to sqrt(7)-2. So you can generate other solutions just be adding an integer to that number.

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