How do you solve sqrt(x^2-28)-1=x?

Aug 5, 2015

$x = \emptyset$

Explanation:

Start by isolating the radical on one side of the equation.

$\sqrt{{x}^{2} - 28} - \textcolor{red}{\cancel{\textcolor{b l a c k}{1}}} + \textcolor{red}{\cancel{\textcolor{b l a c k}{1}}} = x + 1$

Any possible solution to this equation must satisfy the conditions

• ${x}^{2} - 28 \ge 0$ $\to$ for real numbers, you can only take the square root from positive numbers.
• $x - 1 \ge 0 \implies x \ge 1$ $\to$ the square root of a real number can only be a positive number.

In order for the expression under the radical to be positive, you need

${x}^{2} - 28 \ge 0$

${x}^{2} \ge 28$

$| x | \ge \sqrt{28} \implies x \le - \sqrt{28} \vee x \ge \sqrt{28}$

But since $x \ge 1$ is need for the right side of the equation, your overall condition will be $x \ge \sqrt{28}$.

So, square both sides of the equation to get rid of the radical

${\left(\sqrt{{x}^{2} - 28}\right)}^{2} = {\left(x + 1\right)}^{2}$

$\textcolor{red}{\cancel{\textcolor{b l a c k}{{x}^{2}}}} - 28 = \textcolor{red}{\cancel{\textcolor{b l a c k}{{x}^{2}}}} + 2 x + 1$

This is equivalent to

$2 x = - 29 \implies x = \textcolor{red}{\cancel{\textcolor{b l a c k}{- \frac{29}{2}}}}$

As you can see, this is an extraneous solution because it doesn't meet the condition $x \ge \sqrt{28}$. This means that the original equation has no real solutions.