﻿ If x-\sqrt{x} = 13, find the value of x-\dfrac{13}{\sqrt{x} }

Method 1

Given condition

x-\sqrt{x} = 13
(1)

Divide both sides by \sqrt{x}, then

\sqrt{x} -1=\dfrac{13}{\sqrt{x} }

Move the terms with variable to the LHS and constant to the RHS

\sqrt{x}-\dfrac{13}{\sqrt{x} } =1
(2)

Square both sides of (2)

(\sqrt{x})^2-2\cdot 13+(\dfrac{13}{\sqrt{x} } )^2 =1

(\sqrt{x})^2+2\cdot 13+(\dfrac{13}{\sqrt{x} } )^2-4\cdot 13 =1

(\sqrt{x}+\dfrac{13}{\sqrt{x} })^2 =53

Taking the square root of both sides gives

\sqrt{x}+\dfrac{13}{\sqrt{x} } = \sqrt{53}
(3)

Then, solve the system of equaitons of (2) and (3)

\sqrt{x} = \dfrac{\sqrt{53}+1 }{2}

\dfrac{13}{\sqrt{x} } = \dfrac{\sqrt{53}-1 }{2}

Substitution to calculate the value of the expression

x-\dfrac{13}{\sqrt{x} }

= (\dfrac{\sqrt{53}+1 }{2} )^2-\dfrac{\sqrt{53}-1 }{2}

=\dfrac{53+1+2\sqrt{53}-2\sqrt{53}+2 }{4}

=\dfrac{56}{4}

=14

Steven Zheng posted 2 months ago

Method 2

Given x-\sqrt{x} = 13

Substitute 13 to the expression

x-\dfrac{13}{\sqrt{x} }

=x-\dfrac{x-\sqrt{x}}{\sqrt{x} }

=x-\sqrt{x}+1

=13+1=14

Steven Zheng posted 2 months ago

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