Question

Are these steps for finding the solutions of \(\sqrt {x + 3} = 3 - x\)correct?

(1) \(\sqrt {x + 3} = 3 - x\) is given;

(2) \(x + 3 = {x^2} - 6x + 9\), obtained by squaring both sides of (1);

(3) \(0 = {x^2} - 7x + 6\), obtained by subtracting \(x + 3\)from both sides of (2);

(4) \(0 = \left( {x - 1} \right)\left( {x - 6} \right)\), obtained by factoring the right-hand side of (3);

(5) \(x = 1orx = 6\), which follows from (4) because \(ab = 0\) implies that\(a = 0orb = 0\).

Short answer

There is an error in second step, other than that everything is correct.

Step-by-step solution

Introduction\

At the beginning of the step (2), there is \(\sqrt {x + 3} = 3 - x\)

Now, square both sides to obtain the following:

\({\left( {\sqrt {x + 3} } \right)^2} = {\left( {3 - x} \right)^2}\)

\( \Rightarrow \left| {x + 3} \right| = {\left( {3 - x} \right)^2}\)

The function\(f(y) = \sqrt y \)is to be positive square root for\(y \ge 0\)

Therefore the value of\({\left( {\sqrt {x + 3} } \right)^2}\)is\(\left| {x + 3} \right|\)because\(\left| {x + 3} \right| \ge 0\)

But in step (2),\({\left( {\sqrt {x + 3} } \right)^2}\)has been taken as\(\left( {x + 3} \right)\)instead of\(\left| {x + 3} \right|\)

By considering it as\(\left( {x + 3} \right)\), both positive as well as negative values of \({\left( {\sqrt {x + 3} } \right)^2}\)are considered. This is the error in the mentioned method. Other than the step mentioned above, the rest of the method is correct.

Extraneous result calculation

Extraneous result get calculated in this method because both negative as well as positive value of \({\left( {\sqrt {x + 3} } \right)^2}\)are considered. So, in order to eliminate the extraneous result and get the correct result, the final results need to be plugged into the original equation for verification.

The result found in this method are\(x = 1andx = 6\).

Plug in both the values in the original equation:

For\(x = 1\),

\(\sqrt {x + 3} = 3 - x\)

\(\sqrt {1 + 3} = 3 - 1\)

\( \Rightarrow 2 = 2\)

For, x=6

\(\sqrt {x + 3} = 3 - x\)

\(\sqrt {1 + 6} = 3 - 6\)

\( \Rightarrow 3 \ne - 3\)

Therefore, it is seen that using this method, an extraneous result is found as x=6. By putting in the value and eliminating the extraneous result, the correct result is found using this method.