Respuesta :

JJtheHelper

Answer:

There are no extraneous solutions

Reasoning:

An extraneous solution is a solution that isn't valid, it might be imaginary like the square root of a negative number.

first we want to isolate z:

1+sqrt(z)=sqrt(z+5)

^2 all      ^2 all

(1+sqrt(z))(1+sqrt(z))=z+5

expand

1+2sqrt(z)+z=z+5

-1              -z  -z -1

2sqrt(z)=4

/2           /2

sqrt(z)=2

^2 all    ^2 all

z=4

Since there is one solution and it is a real number, there are no extraneous solutions.

konrad509

[tex]D:z\geq 0\wedge z+5\geq 0\\D:z\geq 0\wedge z\geq -5\\D:z\geq0\\\\1+\sqrt z=\sqrt{z+5}\\1+2\sqrt z+z=z+5\\2\sqrt z=4\\\sqrt z=2\\z=4[/tex]

[tex]4\in D[/tex] so there are no extraneous solutions.

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