The equation formed by decreasing each root of `ax^(2)+bx+c=0` by 1 is`2x^(2)+8x+2=0` then a. `a=-b` b. `b=-c` c. `c=-a` d. `b=a+c`

Options

A. `a=-b`
B. `b=-c`
C. `c=-a`
D. `b=a+c`

Answer

To solve the problem step by step, we need to analyze the given equations and their roots.

### Step 1: Understand the given equations
We are given that the equation formed by decreasing each root of \( ax^2 + bx + c = 0 \) by 1 is \( 2x^2 + ax + 2 = 0 \).

### Step 2: Identify the roots of the new equation
Let’s denote the roots of the equation \( 2x^2 + ax + 2 = 0 \) as \( \alpha \) and \( \beta \). According to the problem, the roots of the original equation \( ax^2 + bx + c = 0 \) will be \( \alpha + 1 \) and \( \beta + 1 \).

### Step 3: Use Vieta's formulas for the new equation
From Vieta's formulas, we know:
- The sum of the roots \( \alpha + \beta = -\frac{a}{2} \)
- The product of the roots \( \alpha \beta = \frac{2}{2} = 1 \)

### Step 4: Write equations based on Vieta's formulas
From the above, we can write:
1. \( \alpha + \beta = -\frac{a}{2} \) (Equation 1)
2. \( \alpha \beta = 1 \) (Equation 2)

### Step 5: Analyze the original equation
For the original equation \( ax^2 + bx + c = 0 \), the sum and product of the roots \( \alpha + 1 \) and \( \beta + 1 \) can be expressed as:
- The sum of the roots:
\[ (\alpha + 1) + (\beta + 1) = \alpha + \beta + 2 = -\frac{a}{2} + 2 = -\frac{a}{2} + \frac{4}{2} = -\frac{a - 4}{2} \]
This should equal \( -\frac{b}{a} \), thus:
\[ -\frac{a - 4}{2} = -\frac{b}{a} \implies \frac{a - 4}{2} = \frac{b}{a} \]

### Step 6: Cross-multiply and simplify
Cross-multiplying gives:
\[ a(a - 4) = 2b \implies a^2 - 4a = 2b \quad \text{(Equation 3)} \]

### Step 7: Use the product of the roots for the original equation
The product of the roots \( (\alpha + 1)(\beta + 1) \) can be expressed as:
\[ \alpha \beta + \alpha + \beta + 1 = 1 + (-\frac{a}{2}) + 1 = 2 - \frac{a}{2} \]
This should equal \( \frac{c}{a} \), thus:
\[ 2 - \frac{a}{2} = \frac{c}{a} \]

### Step 8: Cross-multiply and simplify
Cross-multiplying gives:
\[ 2a - \frac{a^2}{2} = c \implies c = 2a - \frac{a^2}{2} \quad \text{(Equation 4)} \]

### Step 9: Relate \( b \) and \( c \)
From Equation 3 and Equation 4, we can express \( b \) and \( c \) in terms of \( a \):
1. From Equation 3: \( b = \frac{a^2 - 4a}{2} \)
2. From Equation 4: \( c = 2a - \frac{a^2}{2} \)

### Step 10: Check the options
Now we can check the options:
- a. \( a = -b \)
- b. \( b = -c \)
- c. \( c = -a \)
- d. \( b = a + c \)

After substituting and simplifying, we find that \( b = -c \) holds true.

### Final Answer
The correct option is **b. \( b = -c \)**.