Answer:
x=7 satisfy the equation, so it is the solution.
x=4 doesn't satisfy the equation so it is extraneous solution.
Step-by-step explanation:
The equation given is:
[tex]\sqrt{x-3}+5=x[/tex]
Adding -5 on both sides
[tex]\sqrt{x-3}=x-5[/tex]
Taking square on both sides
[tex](\sqrt{x-3})^2=(x-5)^2[/tex]
Now solving
[tex]x-3 = x^2 -10x+25\\Arranging\\x^2-10x-x+3+25=0\\x^2-11x+28=0\\Factorizingx^2-7x-4x+28=0\\\\x(x-7)-4(x-7)=0\\(x-4)(x-7)=0\\x-4=0 \,\,and\,\, x-7=0\\x=4 \,\,and\,\, x=7[/tex]
Verifying solutions:
Putting x=4 in the equation
[tex]\sqrt{x-3}+5=x[/tex]
[tex]\sqrt{4-3}+5=4[/tex]
[tex]\sqrt{1}+5=4[/tex]
[tex]1+5=4[/tex]
[tex]6\neq 4[/tex]
So, x=4 doesn't satisfy the equation so it is extraneous solution.
Now Putting x=7 and verifying
[tex]\sqrt{x-3}+5=x[/tex]
[tex]\sqrt{7-3}+5=7[/tex]
[tex]\sqrt{4}+5=7[/tex]
[tex]2+5=7[/tex]
[tex]7=7[/tex]
x=7 satisfy the equation, so it is the solution.
