The equation in the image is as follows:

$\sqrt{x - 2 \sqrt{x - 1}} + \sqrt{x + 3 - 4 \sqrt{x - 1}} = 1$

Let's go step-by-step to solve this equation.

Step 1: Substitution

Let’s make a substitution to simplify the nested square roots: $y = \sqrt{x - 1}$ Thus: $x = y^2 + 1$

Now substitute $x = y^2 + 1$ into the original equation.

1. For the first square root: $\sqrt{x - 2 \sqrt{x - 1}} = \sqrt{(y^2 + 1) - 2y} = \sqrt{y^2 - 2y + 1} = \sqrt{(y - 1)^2} = |y - 1|$

2. For the second square root: $\sqrt{x + 3 - 4 \sqrt{x - 1}} = \sqrt{(y^2 + 1) + 3 - 4y} = \sqrt{y^2 - 4y + 4} = \sqrt{(y - 2)^2} = |y - 2|$

So the equation becomes: $|y - 1| + |y - 2| = 1$

Step 2: Solve the absolute value equation

Now, we solve the equation $|y - 1| + |y - 2| = 1$. To do this, we consider different cases based on the value of $y$.

• Case 1: $y \geq 2$

In this case, both $y - 1$ and $y - 2$ are non-negative, so: $(y - 1) + (y - 2) = 1$ Simplifying: $2y - 3 = 1 \implies 2y = 4 \implies y = 2$ Thus, $y = 2$ is a solution for this case.

• Case 2: $1 \leq y < 2$

In this case, $y - 1$ is non-negative, but $y - 2$ is negative, so: $(y - 1) + -(y - 2) = 1$ Simplifying: $(y - 1) - (y - 2) = 1 \implies 1 = 1$ This is true for all $y$ in the interval $1 \leq y < 2$.

• Case 3: $y < 1$

In this case, both $y - 1$ and $y - 2$ are negative, so: $-(y - 1) + -(y - 2) = 1$ Simplifying: $-2y + 3 = 1 \implies -2y = -2 \implies y = 1$ Thus, $y = 1$ is a solution for this case.

Step 3: Back-substitution

Now we back-substitute the values of $y$ into the original substitution $y = \sqrt{x - 1}$:

• If $y = 2$, then: $2 = \sqrt{x - 1} \implies x - 1 = 4 \implies x = 5$

• If $1 \leq y < 2$, then the equation is satisfied for all $1 \leq y < 2$. Therefore: $1 \leq \sqrt{x - 1} < 2$ Squaring both sides: $1 \leq x - 1 < 4 \implies 2 \leq x < 5$

• If $y = 1$, then: $1 = \sqrt{x - 1} \implies x - 1 = 1 \implies x = 2$

Final Answer:

The solutions to the equation are: $x = 2 \quad \text{and} \quad 2 \leq x \leq 5$

Would you like more details or have any questions about this process?

Related Questions:

1. How do you handle equations with nested radicals?
2. What are the properties of absolute value functions?
3. Can you explain why squaring both sides of an equation can introduce extraneous solutions?
4. What are the common techniques to simplify radical expressions?
5. How would the solution change if the right-hand side of the equation were different, say 2 instead of 1?

Tip:

When solving radical equations, always check for extraneous solutions, especially after squaring both sides, as this process can introduce invalid answers.