Question

The equation $\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+15-8\sqrt{x-1}}=2$ has

• A. No real root
• B. At least one real root
• C. Exactly one root
• D. An infinite number of real roots

Answer 1

The correct option is B

At least one real root

Solve the equation:

The given equation is

$$\begin{gathered} \sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+15-8\sqrt{x-1}}=2 \\ \sqrt{{\left(\sqrt{x-1}-2\right)}^2}+\sqrt{{\left(\sqrt{x-1}-4\right)}^2}=2 \\ \pm \left(\sqrt{x-1}-2\right)\pm \left(\sqrt{x-1}-4\right)=2\left(1\right) \end{gathered}$$

Case 1:

$$\begin{gathered} \left(\sqrt{x-1}-2\right)+\left(\sqrt{x-1}-4\right)=2 \\ 2\sqrt{x-1}-6=2 \\ 2\sqrt{x-1}=8 \\ \sqrt{x-1}=4 \\ x-1=16 \\ x=17 \end{gathered}$$

Case 2:

$$\begin{gathered} -\left(\sqrt{x-1}-2\right)-\left(\sqrt{x-1}-4\right)=2 \\ -2\sqrt{x-1}+6=2 \\ 2\sqrt{x-1}=-4 \\ \sqrt{x-1}=-2 \\ x-1=4 \\ x=5 \end{gathered}$$

For the given equation

$$\begin{gathered} x-1\ge 0 \\ x\ge 1 \end{gathered}$$

Similarly

$$\begin{gathered} x+3-4\sqrt{x-1}\ge 0 \\ {\left(\sqrt{x-1}-2\right)}^2\ge 0 \\ \sqrt{x-1}\ge 2 \\ x-1\ge 4 \\ x\ge 5 \end{gathered}$$

Similarly

$$\begin{gathered} x+15-8\sqrt{x-1}\ge 0 \\ {\left(\sqrt{x-1}-4\right)}^2\ge 0 \\ \sqrt{x-1}\ge 4 \\ x-1\ge 16 \\ x\ge 17 \end{gathered}$$

Therefore the only possible value of $x$ is $17$.

So, it has one real root.

Hence, option $\left(B\right)$ is correct.