Question

Show all work to solve the equation for x. If a solution is extraneous, be sure to identify it in your final answer.

Answer 1

Answer:

There is one extraneous solution at $x = 4$

The only solution is $x=7$

Step-by-step explanation:

Given:

The equation to solve is given as:

$\sqrt{x-3}+5=x$

In order to solve, we will first isolate the radical part.

Adding -5 on both the sides, we get:

$\sqrt{x-3}+5-5=x-5\\\sqrt{x-3}=x-5$

Now, squaring both the sides, we get:

$(\sqrt{x-3})^2=(x-5)^2\\\\x-3=x^2+25-10x\\\\x^2-10x-x+25+3=0\\\\x^2-11x+28=0\\\\(x-4)(x-7)=0\\\\x=4\ or\ x=7$

Now, let us check whether the calculated values satisfy the equations or not.

Plug in 4 and 7 for 'x' and check the value of the equation.

At $x=4$,

$\sqrt{4-3}+5=4\\\sqrt1+5=4\\1+5=4\\6=4$

But, 6 is not equal to 4 and hence 4 is not a solution to the given equation. It is an extraneous solution for the given equation.

Now, at $x=7$

$\sqrt{7-3}+5=7\\\sqrt4+5=7\\2+5=7\\7=7(True)$

Therefore, $x=7$ is a solution to the equation.

Hence, there is one extraneous solution at $x = 4$

The only solution is $x=7$