Question
If x-\dfrac{2}{\sqrt{x} } =5, find the value of x-2\sqrt{x}
Answer 1
Given
x-\dfrac{2}{\sqrt{x} } =5
(1)
Let
a = \sqrt{x}
(2)
then
x = a^2
(3)
Equation (1) is transformed to
a^2-\dfrac{2}{a} = 5
then a cubic equation
a^3-5a -2=0
(4)
Let
A =x-2\sqrt{x}
(5)
Substitute x with a
A = a^2-2a
(6)
or
a^2 = A+2a
(7)
Multiply the equation with a
Aa = a^3-2a^2
(8)
Substitute a^2 with (7)
Aa = a^3-2(A+2a)
Rearrange the terms
a^3-(4+A)a-2A = 0
(9)
Compare the cubic equation with (4)
4+A = 5
Then A = -1
Therefore
x-2\sqrt{x} = -1
Answer 2
x-\dfrac{2}{\sqrt{x} } =5, find the value of x-2\sqrt{x}
x-\dfrac{2}{\sqrt{x} } = 1+4
Rearrange the terms
x-4 = \dfrac{2}{\sqrt{x} } +1
Using difference of squares formula
(\sqrt{x}-2 )(\sqrt{x}+2 ) = \dfrac{2+\sqrt{x} }{\sqrt{x} }
Cancel \sqrt{2} +2 term since it's larger than 1
\sqrt{x}-2 = \dfrac{1}{\sqrt{x} }
(\sqrt{x}-1 )^2 =0
Then x = 1
x-2\sqrt{x} =-1