Question

How do you solve $\sqrt {x + 3} = x - 3$ and find any extraneous solution?

Answer 1

Hint: Here we have an equation with a square root, so we will square both the sides, which will in return give us a quadratic equation.
So, solving the question using the quadratic equation method, we will assume it is in the form $a{x^2} + bx + c = 0$, and use the splitting the middle term formula to find the roots.
After finding the factors, take out the common numbers or variables and factorize the expression. on doing some simplification and we get the required answer.

Complete step-by-step solution:
We have an equation $\sqrt {x + 3} = x - 3$
Since there is a square root present, we know that,
$x + 3 \geqslant 0 \Rightarrow x \geqslant - 3$
When we have real numbers in an equation, the best assumption is that the number in charge is positive, the number whose square root is being taken is positive, that is $x + 3$ must be positive or at the least 0.
$x - 3 \geqslant 0 \Rightarrow x \geqslant 3$
Similarly,
The square root is also assumed to be positive, that is $x - 3$ is positive or at the least 0.
So, any number that satisfies the condition $x \geqslant 3$ can act as the solution for the above equation.
Looking at the equation, $\sqrt {x + 3} = x - 3$
We can remove the square root by squaring both the sides,
$ \Rightarrow {(\sqrt {x + 3} )^2} = {(x - 3)^2}$
Thus, now the equation is,
$ \Rightarrow x + 3 = {x^2} - 6x + 9$
Rearranging to form a quadratic equation,
$ \Rightarrow {x^2} - 7x + 6 = 0$
We now have a quadratic equation ${x^2} - 7x + 6 = 0$
Let us find two factors of $6$ such that when they are added or subtracted, they give us $ - 7$.
If we observe, then two such factors are $ - 1$ and $ - 6$.
$ \Rightarrow (x - 1)(x - 6) = 0$
Now we can write it as $(x - 1) > 3$ and $(x - 6) > 3$
First, we take $x = 1 > 3$ does not satisfy the first condition, hence, it is an extraneous solution.
Therefore, $x = 6$ is the only solution to the given equation.
Checking it,
$ \Rightarrow x = 6$
Putting it in equations,
$ \Rightarrow \sqrt {6 + 3} = 6 - 3$
Adding numbers inside the square root,
$ \Rightarrow \sqrt 9 = 3$
Taking square root,
$ \Rightarrow 3 = 3$
Putting $x = 1$ as solution,
$ \Rightarrow \sqrt {1 + 3} = 1 - 3$
Adding inside the square root,
$ \Rightarrow \sqrt 4 = - 2$
Taking out the square root,
$ \Rightarrow 2 = - 2$
Which is untrue.

Hence, for an equation $\sqrt {x + 3} = x - 3$, the factors are $x = 6$ and $x = 1$ as an extraneous solution.

Note: An extraneous solution is actually a solution that has come out of solving a particular equation, but it is not a valid solution. It is a root of a transformed equation, which does not actually help the equation, but emerges out of the process.
If ${x^2}$ does not have a coefficient, then find two factors of $c$, such that when they are added or subtracted, they give us $b$.
But if ${x^2}$ has a coefficient, then find two factors of $ac$, such that when they are added or subtracted, they give us $b$ .