Prove that the roots of `ax^(2)+2bx + c = 0` will be real distinct if and only if the roots of `(a+c)(ax^(2)+2bx + c)=2(ac-.b^(2))(x^(2)+1)` are imaginary.

Answer

To prove that the roots of the equation \( ax^2 + 2bx + c = 0 \) will be real and distinct if and only if the roots of the equation \( (a+c)(ax^2 + 2bx + c) = 2(ac - b^2)(x^2 + 1) \) are imaginary, we will follow these steps:

### Step 1: Determine the condition for real and distinct roots of the first equation
The roots of the quadratic equation \( ax^2 + 2bx + c = 0 \) are real and distinct if the discriminant \( D \) is greater than 0. The discriminant \( D \) is given by:

\[ D = (2b)^2 - 4ac = 4b^2 - 4ac \]

For the roots to be real and distinct, we require:

\[ D > 0 \implies 4b^2 - 4ac > 0 \implies b^2 - ac > 0 \implies ac - b^2 < 0 \]

### Step 2: Analyze the second equation
The second equation we need to analyze is:

\[ (a+c)(ax^2 + 2bx + c) = 2(ac - b^2)(x^2 + 1) \]

Expanding both sides, we have:

\[ (a+c)(ax^2 + 2bx + c) = a^2x^2 + 2abx + ac + acx^2 + 2bcx + c^2 \]

Combining like terms gives:

\[ (a^2 + ac)x^2 + (2ab + 2bc)x + (ac + c^2) = 2(ac - b^2)(x^2 + 1) \]

Expanding the right-hand side:

\[ 2(ac - b^2)x^2 + 2(ac - b^2) \]

### Step 3: Set the coefficients equal
Equating coefficients from both sides, we get:

1. Coefficient of \( x^2 \):
\[ a^2 + ac = 2(ac - b^2) \]

2. Coefficient of \( x \):
\[ 2ab + 2bc = 0 \]

3. Constant term:
\[ ac + c^2 = 2(ac - b^2) \]

### Step 4: Solve the coefficient equations
From the second equation \( 2ab + 2bc = 0 \), we can factor out \( 2b \):

\[ 2b(a + c) = 0 \]

This gives us two cases:
- \( b = 0 \)
- \( a + c = 0 \)

#### Case 1: \( b = 0 \)
If \( b = 0 \), the first equation simplifies to \( ac < 0 \) for real and distinct roots. The second equation will also yield imaginary roots since the left side becomes \( ac \) and the right side becomes \( 2(-b^2) \).

#### Case 2: \( a + c = 0 \)
If \( a + c = 0 \), then \( c = -a \). Substituting this into the discriminant condition gives:

\[ ac - b^2 < 0 \implies -a^2 - b^2 < 0 \]

This is always true, confirming that the roots are imaginary.

### Step 5: Conclusion
Thus, we have shown that the roots of \( ax^2 + 2bx + c = 0 \) are real and distinct if and only if the roots of \( (a+c)(ax^2 + 2bx + c) = 2(ac - b^2)(x^2 + 1) \) are imaginary.