If x, y and z are real numbers and `x+y+z =7 and xy+yz+xz=10`, then what will be greatest value of x?

Options

A. 2
B. 7
C. `sqrt(29)`
D. 5

Answer

To solve the problem, we start with the given equations:

1. \( x + y + z = 7 \)
2. \( xy + yz + zx = 10 \)

We want to find the greatest value of \( x \).

### Step 1: Express \( y + z \) in terms of \( x \)

From the first equation, we can express \( y + z \) as:
\[ y + z = 7 - x \]

### Step 2: Express \( yz \) in terms of \( x \)

Using the second equation, we can express \( yz \) in terms of \( x \):
\[ yz = 10 - xy - zx \]
Substituting \( z = 7 - x - y \) into the equation gives:
\[ yz = 10 - xy - x(7 - x - y) \]
This simplifies to:
\[ yz = 10 - xy - 7x + x^2 + xy \]
Thus, we have:
\[ yz = 10 - 7x + x^2 \]

### Step 3: Set up a quadratic equation

Now we know:
\[ y + z = 7 - x \quad \text{and} \quad yz = 10 - 7x + x^2 \]
Let \( y \) and \( z \) be the roots of the quadratic equation:
\[ t^2 - (y+z)t + yz = 0 \]
Substituting the values we have:
\[ t^2 - (7 - x)t + (10 - 7x + x^2) = 0 \]

### Step 4: Find the discriminant

For \( y \) and \( z \) to be real numbers, the discriminant of this quadratic must be non-negative:
\[ D = (7 - x)^2 - 4(10 - 7x + x^2) \geq 0 \]
Expanding this:
\[ D = (7 - x)^2 - 40 + 28x - 4x^2 \]
\[ D = 49 - 14x + x^2 - 40 + 28x - 4x^2 \]
\[ D = -3x^2 + 14x + 9 \geq 0 \]

### Step 5: Solve the quadratic inequality

To find the values of \( x \) for which \( D \geq 0 \), we solve:
\[ -3x^2 + 14x + 9 = 0 \]
Using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-14 \pm \sqrt{14^2 - 4(-3)(9)}}{2(-3)} \]
Calculating the discriminant:
\[ 14^2 - 4(-3)(9) = 196 + 108 = 304 \]
Thus,
\[ x = \frac{-14 \pm \sqrt{304}}{-6} \]
Calculating \( \sqrt{304} = 4\sqrt{19} \):
\[ x = \frac{14 \pm 4\sqrt{19}}{6} = \frac{7 \pm 2\sqrt{19}}{3} \]

### Step 6: Find the maximum value

The maximum value occurs at:
\[ x = \frac{7 + 2\sqrt{19}}{3} \]

### Conclusion

Thus, the greatest value of \( x \) is:
\[ \boxed{\frac{7 + 2\sqrt{19}}{3}} \]