If `a lt b lt c lt d`, then the quadratic equation `(x-a) (x-c) +2(x-b) (x-d)=0` has its roots

Options

A. real
B. imaginary
C. one root in (a,c)
D. both roots in (a,c)

Answer

To solve the quadratic equation \((x-a)(x-c) + 2(x-b)(x-d) = 0\) given that \(a < b < c < d\), we will first expand the equation and then analyze the roots.

### Step 1: Expand the equation

We start with the equation:

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

Expanding the first term:

\[ (x-a)(x-c) = x^2 - (a+c)x + ac \]

Expanding the second term:

\[ 2(x-b)(x-d) = 2(x^2 - (b+d)x + bd) = 2x^2 - 2(b+d)x + 2bd \]

Now, combine both expansions:

\[ x^2 - (a+c)x + ac + 2x^2 - 2(b+d)x + 2bd = 0 \]

Combining like terms:

\[ (1 + 2)x^2 - [(a+c) + 2(b+d)]x + (ac + 2bd) = 0 \]

This simplifies to:

\[ 3x^2 - (a+c + 2b + 2d)x + (ac + 2bd) = 0 \]

### Step 2: Identify coefficients

From the standard form \(Ax^2 + Bx + C = 0\), we have:

- \(A = 3\)
- \(B = -(a+c + 2b + 2d)\)
- \(C = ac + 2bd\)

### Step 3: Calculate the discriminant

The discriminant \(D\) of a quadratic equation \(Ax^2 + Bx + C = 0\) is given by:

\[ D = B^2 - 4AC \]

Substituting our coefficients:

\[ D = [-(a+c + 2b + 2d)]^2 - 4 \cdot 3 \cdot (ac + 2bd) \]

This simplifies to:

\[ D = (a+c + 2b + 2d)^2 - 12(ac + 2bd) \]

### Step 4: Analyze the discriminant

To determine the nature of the roots, we need to analyze the discriminant \(D\):

1. If \(D > 0\), the equation has two distinct real roots.
2. If \(D = 0\), the equation has exactly one real root (a repeated root).
3. If \(D < 0\), the equation has no real roots (imaginary roots).

Given that \(a < b < c < d\), we can analyze the expression:

- The term \((a+c + 2b + 2d)^2\) is always positive.
- The term \(12(ac + 2bd)\) can be analyzed based on the values of \(a, b, c, d\).

### Step 5: Conclusion about the roots

Since \(a < b < c < d\), it can be inferred that:

- At least one root lies between \(a\) and \(c\) because the function changes sign between these points.
- At least one root lies between \(c\) and \(d\) for the same reason.

Thus, we conclude that the quadratic equation has:

- One root between \(a\) and \(c\).
- One root between \(c\) and \(d\).

### Final Answer

The roots of the quadratic equation \((x-a)(x-c) + 2(x-b)(x-d) = 0\) are such that:

- One root is between \(a\) and \(c\).
- The other root is between \(c\) and \(d\).