No extraneous solutions to this equation.
The equation is
[tex](5x- 4)^{1/2}[/tex] - x = 0
To find the value of x
[tex](5x- 4)^{1/2}[/tex] = x
Now square on both sides of the equation.
We get
5x - 4 =[tex]x^{2}[/tex]
[tex]x^{2}[/tex] - 5x + 4= 0
[tex]x^{2}[/tex] - 4x - x + 4 =0
x( x -4 ) - 1 ( x - 4) =0
( x- 4 ) ( x - 1)= 0
x = 4, 1
We get the value of x as 4 and 1
Now we have to put the value of x in the equation
First, put 4
[tex](5x - 4)^{1/2}[/tex] -x =0
[tex](5 * 4 - 4)^{1/2}[/tex] - 4= 0
[tex](20 - 4)^{1/2}[/tex] - 4 =0
[tex]16^{1/2}[/tex] - 4 = 0
4-4 =0
0 = 0
Now put 1
[tex](5 * 1 - 4)^{1/2}[/tex] - 1 = 0
[tex](5 -4)^{1/2}[/tex] - 1=0
1 - 1=0
0=0
Hence we get that both the value of x is not an extraneous solution.
To know more about the Extraneous solution refer to the link given below:
https://brainly.com/question/2959656
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