Question

How do you solve the equation $\sqrt {x - 1} = x - 3$?

Answer 1

This is a square root function. Find out its domain first and then square on both sides of the equation to convert it in the form of a quadratic equation. Solve the quadratic equation by taking its factors to zero and then compare the solutions to the domain of the function.

Complete step by step answer:
According to the question, we have to show the process to solve the given equation.
$\sqrt {x - 1} = x - 3{\text{ }}.....{\text{(1)}}$
Before solving it, we’ll consider its domain because the function is square root. We know that the quantities in the square root must be non-negative for a function to be defined. So we have:
$\ x - 1 \geqslant 0 \\\ \Rightarrow x \geqslant 1 \\\ \Rightarrow x \in \left[ {1,\infty } \right)$
Thus while solving the equation we have to keep in mind that $x$ shouldn’t be less than 1.
Now, for solving it, let’s take square on both sides of equation (1), we’ll get:
$x - 1 = {\left( {x - 3} \right)^2}$
Now we will apply the formula ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$ on the right hand side, we’ll get:

$$\ x - 1 = {x^2} + 9 - 6x \\\ \Rightarrow {x^2} - 7x + 10 = 0$$

Now factoring the quadratic equation by splitting the middle term into two terms, we’ll get:

$$\ {x^2} - 5x - 2x + 10 = 0 \\\ \Rightarrow x\left( {x - 5} \right) - 2\left( {x - 5} \right) = 0 \\\ \Rightarrow \left( {x - 2} \right)\left( {x - 5} \right) = 0$$

Putting both factors to zero separately, we’ll get:
$\left( {x - 2} \right) = 0{\text{ or }}\left( {x - 5} \right) = 0 \\\ \Rightarrow x = 2{\text{ or }}x = 5$
Both $x = 2$ and $x = 5$ are greater than 1, which was our initial condition, so both of them are valid.

Thus, $x = 2$ and $x = 5$ are the solutions of the equation.

Note: If we are facing any difficulty solving a quadratic equation using factorization method, we can also use a direct formula to find its roots. Let the quadratic equation be:
$y = a{x^2} + bx + c$
The formula to determine its roots is:
$\Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Using this formula we can also find the roots if they are imaginary, complex or irrational numbers.