If a,b,c `in Q`, then roots of the equation `(b+c-2a) x^(2)+(c+a-2b) x+(a+b-2c)=0` are

Options

A. rational
B. irrational
C. non-real
D. equal

Answer

To determine the nature of the roots of the quadratic equation \((b+c-2a)x^2 + (c+a-2b)x + (a+b-2c) = 0\), we will use the discriminant method. The discriminant \(D\) of a quadratic equation \(ax^2 + bx + c = 0\) is given by:

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

### Step 1: Identify coefficients
From the given quadratic equation, we can identify:
- \(A = b + c - 2a\)
- \(B = c + a - 2b\)
- \(C = a + b - 2c\)

### Step 2: Calculate the discriminant
Now, we will calculate the discriminant \(D\):

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

Substituting the values of \(A\), \(B\), and \(C\):

\[ D = (c + a - 2b)^2 - 4(b + c - 2a)(a + b - 2c) \]

### Step 3: Expand \(B^2\)
First, we will expand \(B^2\):

\[ B^2 = (c + a - 2b)^2 = c^2 + a^2 + 4b^2 + 2ac - 4ab - 4bc \]

### Step 4: Expand \(4AC\)
Next, we will expand \(4AC\):

\[ 4AC = 4(b + c - 2a)(a + b - 2c) \]

Using the distributive property, we can expand this expression:

\[ = 4[(b)(a) + (b)(b) - 2(b)(c) + (c)(a) + (c)(b) - 2(c)(c) - 2a(a) - 2a(b) + 4a(c)] \]

This will yield a more complex polynomial, but we will focus on simplifying it later.

### Step 5: Combine and simplify
Now, we will combine \(B^2\) and \(4AC\) to find \(D\):

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

After simplification, we will check if \(D\) is a perfect square or not.

### Step 6: Analyze \(D\)
If \(D\) is a perfect square, the roots are rational. If \(D > 0\) but not a perfect square, the roots are irrational. If \(D = 0\), the roots are equal. If \(D < 0\), the roots are non-real.

### Conclusion
After performing the calculations, we find that \(D\) can be expressed as:

\[ D = 9(a - c)^2 \]

Since \(a, b, c \in \mathbb{Q}\), it follows that \(D\) is a perfect square. Therefore, the roots of the quadratic equation are rational.

### Final Answer
The roots of the equation \((b+c-2a)x^2 + (c+a-2b)x + (a+b-2c) = 0\) are rational.

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