Solve the equations:
Q. `sqrt(x^(2)-3x)=4x^(2)-12x-3`.

Answer

To solve the equation \( \sqrt{x^2 - 3x} = 4x^2 - 12x - 3 \), we will follow these steps:

### Step 1: Square both sides
We start by squaring both sides to eliminate the square root:

\[ \left(\sqrt{x^2 - 3x}\right)^2 = \left(4x^2 - 12x - 3\right)^2 \]

This simplifies to:

\[ x^2 - 3x = (4x^2 - 12x - 3)^2 \]

### Step 2: Set up the equation
Next, we will rearrange the equation to set it to zero:

\[ x^2 - 3x - (4x^2 - 12x - 3)^2 = 0 \]

### Step 3: Substitute \( y \)
Let \( y = x^2 - 3x \). Then we can rewrite the equation as:

\[ \sqrt{y} = 4y - 3 \]

### Step 4: Square both sides again
Squaring both sides again gives:

\[ y = (4y - 3)^2 \]

### Step 5: Expand the right-hand side
Expanding the right-hand side:

\[ y = 16y^2 - 24y + 9 \]

### Step 6: Rearrange to form a quadratic equation
Rearranging gives us:

\[ 16y^2 - 25y + 9 = 0 \]

### Step 7: Factor the quadratic equation
Now we will factor the quadratic equation:

\[ (16y - 9)(y - 1) = 0 \]

### Step 8: Solve for \( y \)
Setting each factor to zero gives:

1. \( 16y - 9 = 0 \) which gives \( y = \frac{9}{16} \)
2. \( y - 1 = 0 \) which gives \( y = 1 \)

### Step 9: Substitute back to find \( x \)
Recall that \( y = x^2 - 3x \). We will substitute back to find \( x \):

1. For \( y = 1 \):
\[ x^2 - 3x - 1 = 0 \]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{3 \pm \sqrt{9 + 4}}{2} = \frac{3 \pm \sqrt{13}}{2} \]

2. For \( y = \frac{9}{16} \):
\[ x^2 - 3x - \frac{9}{16} = 0 \]
Again using the quadratic formula:
\[ x = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot \left(-\frac{9}{16}\right)}}{2 \cdot 1} = \frac{3 \pm \sqrt{9 + \frac{36}{16}}}{2} = \frac{3 \pm \sqrt{\frac{144}{16}}}{2} = \frac{3 \pm 3}{2} \]
This gives \( x = 3 \) or \( x = 0 \).

### Step 10: Final solutions
Thus, the solutions for \( x \) are:

\[ x = \frac{3 + \sqrt{13}}{2}, \quad x = \frac{3 - \sqrt{13}}{2}, \quad x = 3, \quad x = 0 \]