Solve the following quations by using qardratic formula:
`(1)/(x-2)+(1)/(x-3)+(1)/(x-4)=0`

Answer

To solve the equation \(\frac{1}{x-2} + \frac{1}{x-3} + \frac{1}{x-4} = 0\) using the quadratic formula, we will follow these steps:

### Step 1: Find a common denominator
The common denominator for the fractions is \((x-2)(x-3)(x-4)\). We can rewrite the equation as:

\[ \frac{(x-3)(x-4) + (x-2)(x-4) + (x-2)(x-3)}{(x-2)(x-3)(x-4)} = 0 \]

Since the denominator cannot be zero, we can focus on setting the numerator equal to zero:

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

### Step 2: Expand the numerator
Now, we will expand each term in the numerator:

1. \((x-3)(x-4) = x^2 - 7x + 12\)
2. \((x-2)(x-4) = x^2 - 6x + 8\)
3. \((x-2)(x-3) = x^2 - 5x + 6\)

Adding these together:

\[ (x^2 - 7x + 12) + (x^2 - 6x + 8) + (x^2 - 5x + 6) = 0 \]

Combining like terms gives:

\[ 3x^2 - 18x + 26 = 0 \]

### Step 3: Identify coefficients
From the equation \(3x^2 - 18x + 26 = 0\), we identify \(a = 3\), \(b = -18\), and \(c = 26\).

### Step 4: Apply the quadratic formula
The quadratic formula is given by:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Substituting in our values:

\[ x = \frac{-(-18) \pm \sqrt{(-18)^2 - 4 \cdot 3 \cdot 26}}{2 \cdot 3} \]

### Step 5: Calculate the discriminant
Calculating \(b^2 - 4ac\):

\[ (-18)^2 = 324 \]
\[ 4 \cdot 3 \cdot 26 = 312 \]
\[ b^2 - 4ac = 324 - 312 = 12 \]

### Step 6: Substitute back into the formula
Now substituting back into the quadratic formula:

\[ x = \frac{18 \pm \sqrt{12}}{6} \]

### Step 7: Simplify the square root
We can simplify \(\sqrt{12}\) as follows:

\[ \sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3} \]

So now we have:

\[ x = \frac{18 \pm 2\sqrt{3}}{6} \]

### Step 8: Simplify the expression
Dividing each term by 6:

\[ x = \frac{18}{6} \pm \frac{2\sqrt{3}}{6} \]
\[ x = 3 \pm \frac{\sqrt{3}}{3} \]

### Final Solutions
Thus, the solutions are:

\[ x = 3 + \frac{\sqrt{3}}{3} \quad \text{and} \quad x = 3 - \frac{\sqrt{3}}{3} \]