Answer :
To solve the equation [tex]\(\sqrt{x+5} = |2x|\)[/tex], let's break it down step by step. We'll consider the absolute value function's piecewise nature and solve accordingly.
### Step 1: Understand the absolute value function
We know that [tex]\(|2x|\)[/tex] is defined as:
[tex]\[ |2x| = \begin{cases} 2x & \text{if } x \geq 0 \\ -2x & \text{if } x < 0 \end{cases} \][/tex]
### Step 2: Case analysis
#### Case 1: [tex]\(x \geq 0\)[/tex]
For [tex]\(x \geq 0\)[/tex], [tex]\(|2x| = 2x\)[/tex]. Hence, the equation becomes:
[tex]\[ \sqrt{x + 5} = 2x \][/tex]
Square both sides to eliminate the square root:
[tex]\[ (\sqrt{x + 5})^2 = (2x)^2 \implies x + 5 = 4x^2 \][/tex]
Rearrange the equation:
[tex]\[ 4x^2 - x - 5 = 0 \][/tex]
Solve this quadratic equation using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex]:
[tex]\[ a = 4, \quad b = -1, \quad c = -5 \][/tex]
[tex]\[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(4)(-5)}}{2(4)} = \frac{1 \pm \sqrt{1 + 80}}{8} = \frac{1 \pm \sqrt{81}}{8} = \frac{1 \pm 9}{8} \][/tex]
Thus, the solutions are:
[tex]\[ x = \frac{10}{8} = \frac{5}{4}, \quad x = \frac{-8}{8} = -1 \][/tex]
Since we are considering [tex]\(x \geq 0\)[/tex] in this case, we only take the positive solution:
[tex]\[ x = \frac{5}{4} \][/tex]
#### Case 2: [tex]\(x < 0\)[/tex]
For [tex]\(x < 0\)[/tex], [tex]\(|2x| = -2x\)[/tex]. Hence, the equation becomes:
[tex]\[ \sqrt{x + 5} = -2x \][/tex]
Square both sides to eliminate the square root:
[tex]\[ (\sqrt{x + 5})^2 = (-2x)^2 \implies x + 5 = 4x^2 \][/tex]
This results in the same quadratic equation as in Case 1:
[tex]\[ 4x^2 - x - 5 = 0 \][/tex]
The solutions are again:
[tex]\[ x = \frac{5}{4}, \quad x = -1 \][/tex]
Since we are considering [tex]\(x < 0\)[/tex] in this case, we only take the negative solution:
[tex]\[ x = -1 \][/tex]
### Step 3: Verify solutions in the original equation
We have found two candidate solutions, [tex]\(x = \frac{5}{4}\)[/tex] and [tex]\(x = -1\)[/tex].
- For [tex]\(x = \frac{5}{4}\)[/tex]:
[tex]\[ \sqrt{\frac{5}{4} + 5} = \sqrt{\frac{5}{4} + \frac{20}{4}} = \sqrt{\frac{25}{4}} = \frac{5}{2} \][/tex]
[tex]\[ |2 \cdot \frac{5}{4}| = | \frac{5}{2}| = \frac{5}{2} \][/tex]
- For [tex]\(x = -1\)[/tex]:
[tex]\[ \sqrt{-1 + 5} = \sqrt{4} = 2 \][/tex]
[tex]\[ |2 \cdot (-1)| = |-2| = 2 \][/tex]
Both solutions satisfy the original equation.
### Conclusion
There are [tex]\(\boxed{2}\)[/tex] real solutions to the equation [tex]\(\sqrt{x + 5} = |2x|\)[/tex]. The solutions are [tex]\(x = \frac{5}{4}\)[/tex] and [tex]\(x = -1\)[/tex].
### Step 1: Understand the absolute value function
We know that [tex]\(|2x|\)[/tex] is defined as:
[tex]\[ |2x| = \begin{cases} 2x & \text{if } x \geq 0 \\ -2x & \text{if } x < 0 \end{cases} \][/tex]
### Step 2: Case analysis
#### Case 1: [tex]\(x \geq 0\)[/tex]
For [tex]\(x \geq 0\)[/tex], [tex]\(|2x| = 2x\)[/tex]. Hence, the equation becomes:
[tex]\[ \sqrt{x + 5} = 2x \][/tex]
Square both sides to eliminate the square root:
[tex]\[ (\sqrt{x + 5})^2 = (2x)^2 \implies x + 5 = 4x^2 \][/tex]
Rearrange the equation:
[tex]\[ 4x^2 - x - 5 = 0 \][/tex]
Solve this quadratic equation using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex]:
[tex]\[ a = 4, \quad b = -1, \quad c = -5 \][/tex]
[tex]\[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(4)(-5)}}{2(4)} = \frac{1 \pm \sqrt{1 + 80}}{8} = \frac{1 \pm \sqrt{81}}{8} = \frac{1 \pm 9}{8} \][/tex]
Thus, the solutions are:
[tex]\[ x = \frac{10}{8} = \frac{5}{4}, \quad x = \frac{-8}{8} = -1 \][/tex]
Since we are considering [tex]\(x \geq 0\)[/tex] in this case, we only take the positive solution:
[tex]\[ x = \frac{5}{4} \][/tex]
#### Case 2: [tex]\(x < 0\)[/tex]
For [tex]\(x < 0\)[/tex], [tex]\(|2x| = -2x\)[/tex]. Hence, the equation becomes:
[tex]\[ \sqrt{x + 5} = -2x \][/tex]
Square both sides to eliminate the square root:
[tex]\[ (\sqrt{x + 5})^2 = (-2x)^2 \implies x + 5 = 4x^2 \][/tex]
This results in the same quadratic equation as in Case 1:
[tex]\[ 4x^2 - x - 5 = 0 \][/tex]
The solutions are again:
[tex]\[ x = \frac{5}{4}, \quad x = -1 \][/tex]
Since we are considering [tex]\(x < 0\)[/tex] in this case, we only take the negative solution:
[tex]\[ x = -1 \][/tex]
### Step 3: Verify solutions in the original equation
We have found two candidate solutions, [tex]\(x = \frac{5}{4}\)[/tex] and [tex]\(x = -1\)[/tex].
- For [tex]\(x = \frac{5}{4}\)[/tex]:
[tex]\[ \sqrt{\frac{5}{4} + 5} = \sqrt{\frac{5}{4} + \frac{20}{4}} = \sqrt{\frac{25}{4}} = \frac{5}{2} \][/tex]
[tex]\[ |2 \cdot \frac{5}{4}| = | \frac{5}{2}| = \frac{5}{2} \][/tex]
- For [tex]\(x = -1\)[/tex]:
[tex]\[ \sqrt{-1 + 5} = \sqrt{4} = 2 \][/tex]
[tex]\[ |2 \cdot (-1)| = |-2| = 2 \][/tex]
Both solutions satisfy the original equation.
### Conclusion
There are [tex]\(\boxed{2}\)[/tex] real solutions to the equation [tex]\(\sqrt{x + 5} = |2x|\)[/tex]. The solutions are [tex]\(x = \frac{5}{4}\)[/tex] and [tex]\(x = -1\)[/tex].
