If the equation `2x^(2)-7x+9=0` and `ax^(2)+bx+18=0` have a common root, then `(a, b inR)`

Options

A. `a=2, b=-7`
B. `a=(-7)/(2), b=1`
C. `a=4, b=-14`
D. `a=4, b=-7`

Answer

To solve the problem, we need to find the values of \( a \) and \( b \) such that the equations \( 2x^2 - 7x + 9 = 0 \) and \( ax^2 + bx + 18 = 0 \) have a common root.

### Step 1: Find the roots of the first equation
The first equation is given as:
\[ 2x^2 - 7x + 9 = 0 \]
We will use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 2 \), \( b = -7 \), and \( c = 9 \).

Calculating the discriminant:
\[ b^2 - 4ac = (-7)^2 - 4 \cdot 2 \cdot 9 = 49 - 72 = -23 \]
Since the discriminant is negative, the roots are complex.

Now, substituting into the quadratic formula:
\[ x = \frac{7 \pm \sqrt{-23}}{4} = \frac{7 \pm i\sqrt{23}}{4} \]
Thus, the roots are:
\[ x_1 = \frac{7 + i\sqrt{23}}{4}, \quad x_2 = \frac{7 - i\sqrt{23}}{4} \]

### Step 2: Set up the condition for a common root
Let’s denote one of the roots (for example, \( x_1 \)) as a common root for the second equation:
\[ ax^2 + bx + 18 = 0 \]
Substituting \( x_1 \) into the second equation:
\[ a\left(\frac{7 + i\sqrt{23}}{4}\right)^2 + b\left(\frac{7 + i\sqrt{23}}{4}\right) + 18 = 0 \]

### Step 3: Use the condition of common roots
Since both equations have a common root, we can use the property of the coefficients. The ratios of the coefficients must be equal:
\[ \frac{2}{a} = \frac{-7}{b} = \frac{9}{18} \]

### Step 4: Solve for \( a \) and \( b \)
From the ratio \( \frac{9}{18} = \frac{1}{2} \):
\[ \frac{2}{a} = \frac{1}{2} \implies 2 \cdot 2 = a \implies a = 4 \]

Now, using the second ratio:
\[ \frac{-7}{b} = \frac{1}{2} \implies -7 \cdot 2 = b \implies b = -14 \]

### Final Answer
Thus, the values of \( a \) and \( b \) are:
\[ (a, b) = (4, -14) \]