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Which statement best reflects the solution(s) of the equation? √2x−1−x+2=0

answer choices:

There is only one solution: x = 1. The solution x = 5 is an extraneous solution.

There are two solutions: x = 1 and x = 5.

 There is only one solution: x = 5. The solution x = 1 is an extraneous solution.

There is only one solution: x = 5. The solution x = 0 is an extraneous solution.

Respuesta :

W0lf93
Simplifing the expression and moving to the RHS we have âš2x = x +1 Let's check for extraneous solution by squaring both sides. I assume the x and 2 are together. So we have 2x = (x+1)^2 This gives 2x = x^2 + 2x +1. So x^2 + 2x +1 -2x = 0. Then we have x^2 =-1. When x =1; 1 is not equal to -1 so x =1 is an extraneous solution.
kudzordzifrancis

Answer:

There is only one solution: x = 5. The solution x = 1 is an extraneous solution.

Step-by-step explanation:

The given expression is

[tex]\sqrt{2x-1} -x+2=0[/tex]


We rewrite the expression so that we equate everything to zero to obtain,

[tex]\sqrt{2x-1} =x-2[/tex]


We now square both sides to obtain,

[tex](\sqrt{2x-1})^2 =(x-2)^2[/tex]


This simplifies to give us,

[tex]2x-1=(x-2)^2[/tex]


We now expand the parenthesis to get,

[tex]2x-1=x^2-4x+4[/tex]


We now group the terms so that we can obtain the quadratic equation;

[tex]0=x^2-4x-2x+4+1[/tex]


[tex]0=x^2-6x+5[/tex]

[tex]x^2-6x+5=0[/tex]


We split the middle term to obtain,


[tex]x^2-x-5x+5=0[/tex]


We factor to obtain,

[tex]x(x-1)-5(x-1)=0[/tex]


[tex](x-1)(x-5)=0[/tex]


[tex]\Rightarrow (x-1)=0\:or\:(x-5)=0[/tex]


[tex]\Rightarrow x=1\:or\:x=5[/tex]


Let us check to see whether  the two values satisfy the given equation.


When [tex]x=1[/tex], we get,

[tex]\sqrt{2(1)-1} -1+2=0[/tex]

[tex]\Rightarrow \sqrt{1} +1=0[/tex]


[tex]\Rightarrow 1 +1=0[/tex]


[tex]\Rightarrow 2=0[/tex]


This statement is false. Hence [tex]x=1[/tex] is an extraneous solution.


When [tex]x=5[/tex], we get,

[tex]\sqrt{2(5)-1} -5+2=0[/tex]

[tex]\Rightarrow \sqrt{9} -3=0[/tex]


[tex]\Rightarrow 3-3=0[/tex]


[tex]\Rightarrow 0=0[/tex]


This statement is very trues.


Therefore [tex]x=5[/tex] is the only solution.


There is only one solution: x = 5. The solution x = 1 is an extraneous solution.
















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