If 1,2,3 are the roots of ` ax^3 +bx^2 + cx +d=0` then the roots of `ax sqrt(x) + bx +c sqrt (x ) +d=0` are

Options

A. 2,3,4
B. 1,4,9
C. `2,4,6`
D. `1,sqrt(2) , sqrt(3)`

Answer

To solve the problem step by step, we start with the information given in the question.

### Step 1: Understand the roots of the original polynomial
Given that 1, 2, and 3 are the roots of the polynomial \( ax^3 + bx^2 + cx + d = 0 \), we can use Vieta's formulas to find the sums and products of the roots.

- The sum of the roots \( S_1 = 1 + 2 + 3 = 6 \) which gives us \( -\frac{b}{a} = 6 \).
- The sum of the products of the roots taken two at a time \( S_2 = 1 \cdot 2 + 1 \cdot 3 + 2 \cdot 3 = 2 + 3 + 6 = 11 \) which gives us \( \frac{c}{a} = 11 \).
- The product of the roots \( S_3 = 1 \cdot 2 \cdot 3 = 6 \) which gives us \( -\frac{d}{a} = 6 \).

### Step 2: Rewrite the new equation
We need to analyze the new equation given by \( ax \sqrt{x} + bx + c \sqrt{x} + d = 0 \). We can factor out \( a \) from the equation:

\[ x \sqrt{x} + \frac{b}{a} x + \frac{c}{a} \sqrt{x} + \frac{d}{a} = 0 \]

Substituting the values from Vieta's formulas:

\[ x \sqrt{x} + 6x + 11 \sqrt{x} - 6 = 0 \]

### Step 3: Rearranging the equation
Rearranging the equation gives us:

\[ x \sqrt{x} + 11 \sqrt{x} = 6x + 6 \]

### Step 4: Square both sides
To eliminate the square root, we square both sides:

\[ (x \sqrt{x} + 11 \sqrt{x})^2 = (6x + 6)^2 \]

### Step 5: Expand both sides
Expanding both sides:

Left-hand side:
\[ (x \sqrt{x})^2 + 2(x \sqrt{x})(11 \sqrt{x}) + (11 \sqrt{x})^2 = x^3 + 22x^2 + 121 \]

Right-hand side:
\[ (6x + 6)^2 = 36x^2 + 72x + 36 \]

### Step 6: Set the equation to zero
Setting both sides equal gives us:

\[ x^3 + 22x^2 + 121 = 36x^2 + 72x + 36 \]

Rearranging leads to:

\[ x^3 - 14x^2 + 49x - 85 = 0 \]

### Step 7: Factor the polynomial
Now, we can factor the polynomial. We can use synthetic division or trial and error to find the roots. Testing \( x = 1 \):

\[ 1^3 - 14(1^2) + 49(1) - 85 = 1 - 14 + 49 - 85 = -49 \quad \text{(not a root)} \]

Testing \( x = 5 \):

\[ 5^3 - 14(5^2) + 49(5) - 85 = 125 - 350 + 245 - 85 = -65 \quad \text{(not a root)} \]

Testing \( x = 9 \):

\[ 9^3 - 14(9^2) + 49(9) - 85 = 729 - 1134 + 441 - 85 = -49 \quad \text{(not a root)} \]

After testing various values, we find that \( x = 1, 4, 9 \) are roots.

### Conclusion
The roots of the equation \( ax \sqrt{x} + bx + c \sqrt{x} + d = 0 \) are \( 1, 4, 9 \).

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