If `3x^(2) -17x+10 =0 and x^(2)-5x+m =0` has a common root, then sum of all possible real values of 'm' is:

Options

A. 0
B. `-26/9`
C. `29/9`
D. `26/3`

Answer

To solve the problem, we need to find the values of \( m \) for which the quadratic equations \( 3x^2 - 17x + 10 = 0 \) and \( x^2 - 5x + m = 0 \) have a common root. We will follow these steps:

### Step 1: Find the roots of the first quadratic equation
The first equation is:
\[ 3x^2 - 17x + 10 = 0 \]
We can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 3 \), \( b = -17 \), and \( c = 10 \).

Calculating the discriminant:
\[ b^2 - 4ac = (-17)^2 - 4 \cdot 3 \cdot 10 = 289 - 120 = 169 \]
Now, substituting into the quadratic formula:
\[ x = \frac{17 \pm \sqrt{169}}{2 \cdot 3} = \frac{17 \pm 13}{6} \]
Calculating the two roots:
\[ x_1 = \frac{30}{6} = 5 \quad \text{and} \quad x_2 = \frac{4}{6} = \frac{2}{3} \]

### Step 2: Use the common root in the second quadratic equation
The second equation is:
\[ x^2 - 5x + m = 0 \]
We will substitute both roots \( x_1 = 5 \) and \( x_2 = \frac{2}{3} \) into this equation to find the corresponding values of \( m \).

#### Substituting \( x = 5 \):
\[ 5^2 - 5 \cdot 5 + m = 0 \]
\[ 25 - 25 + m = 0 \implies m = 0 \]

#### Substituting \( x = \frac{2}{3} \):
\[ \left(\frac{2}{3}\right)^2 - 5 \cdot \frac{2}{3} + m = 0 \]
Calculating:
\[ \frac{4}{9} - \frac{10}{3} + m = 0 \]
Converting \( \frac{10}{3} \) to a fraction with a denominator of 9:
\[ \frac{10}{3} = \frac{30}{9} \]
So:
\[ \frac{4}{9} - \frac{30}{9} + m = 0 \implies -\frac{26}{9} + m = 0 \implies m = \frac{26}{9} \]

### Step 3: Sum of all possible values of \( m \)
The possible values of \( m \) are \( 0 \) and \( \frac{26}{9} \). Therefore, the sum of all possible values of \( m \) is:
\[ 0 + \frac{26}{9} = \frac{26}{9} \]

### Final Answer:
The sum of all possible real values of \( m \) is:
\[ \frac{26}{9} \]

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