The equation whose roots are 'K' times the roots of the equation `ax^(2)+bx+c=0` is

Options

A. `ax^(2)+Kbx+Kc=0`
B. `ax^(2)+K^(2)by+K^(2)c=0`
C. `ax^(2)+Kbx+K^(2)c=0`
D. `ax^(2)+K^(2)bx+Kc=0`

Answer

To find the equation whose roots are 'K' times the roots of the equation \( ax^2 + bx + c = 0 \), we can follow these steps:

### Step 1: Identify the roots of the original equation
Let the roots of the equation \( ax^2 + bx + c = 0 \) be \( \alpha \) and \( \beta \).

### Step 2: Use Vieta's formulas
From Vieta's formulas, we know:
- The sum of the roots \( \alpha + \beta = -\frac{b}{a} \)
- The product of the roots \( \alpha \beta = \frac{c}{a} \)

### Step 3: Determine the new roots
The new roots we are looking for are \( K\alpha \) and \( K\beta \).

### Step 4: Write the new equation
The general form of a quadratic equation with roots \( r_1 \) and \( r_2 \) is given by:
\[ x^2 - (r_1 + r_2)x + r_1r_2 = 0 \]
Substituting \( r_1 = K\alpha \) and \( r_2 = K\beta \):
\[ x^2 - (K\alpha + K\beta)x + (K\alpha)(K\beta) = 0 \]

### Step 5: Simplify the equation
Now, substituting the values of \( \alpha + \beta \) and \( \alpha \beta \):
\[ x^2 - K(\alpha + \beta)x + K^2(\alpha \beta) = 0 \]
Substituting Vieta's results:
\[ x^2 - K\left(-\frac{b}{a}\right)x + K^2\left(\frac{c}{a}\right) = 0 \]

### Step 6: Rearranging the equation
This simplifies to:
\[ x^2 + \frac{Kb}{a}x + \frac{K^2c}{a} = 0 \]

### Step 7: Multiply through by \( a \)
To eliminate the fraction, multiply the entire equation by \( a \):
\[ ax^2 + Kbx + K^2c = 0 \]

### Final Result
Thus, the equation whose roots are \( K \) times the roots of the equation \( ax^2 + bx + c = 0 \) is:
\[ ax^2 + Kbx + K^2c = 0 \]

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