Given:
The equation is:
[tex]\sqrt{x+7}+5=x[/tex]
To find:
Whether [tex]x=2[/tex] and [tex]x=9[/tex] both are solutions or one/both of them extraneous solutions.
Solution:
We have,
[tex]\sqrt{x+7}+5=x[/tex]
Subtract 5 from both sides.
[tex]\sqrt{x+7}=x-5[/tex]
Taking square on both sides, we get
[tex]x+7=(x-5)^2[/tex]
[tex]x+7=x^2-10x+25[/tex]
[tex]0=x^2-10x+25-x-7[/tex]
[tex]0=x^2-11x+18[/tex]
Splitting the middle term, we get
[tex]x^2-2x-9x+18=0[/tex]
[tex]x(x-2)-9(x-2)=0[/tex]
[tex](x-2)(x-9)=0[/tex]
[tex]x=2,9[/tex]
Now, substitute [tex]x=2[/tex] in the given equation.
[tex]\sqrt{2+7}+5=2[/tex]
[tex]\sqrt{9}+5=2[/tex]
[tex]3+5=2[/tex]
[tex]8=2[/tex]
This statement is false because [tex]8\neq 2[/tex]. So, 2 is an extraneous solution.
Substitute [tex]x=9[/tex] in the given equation.
[tex]\sqrt{9+7}+5=9[/tex]
[tex]\sqrt{16}+5=9[/tex]
[tex]4+5=9[/tex]
[tex]9=9[/tex]
This statement is true. So, 9 is a solution of given equation.
Therefore, 2 is an extraneous solution and 9 is a solution of given equation.
