The equation whose roots are smaller by 1 then those of `2x ^2-5x +6=0` is

Options

A. `2x^(2) - 9x + 13 = 0`
B. `2x^(2) - x + 3 = 0`
C. `2x^(2) + 9x + 13 = 0`
D. `2x^(2) +x + 3 = 0`

Answer

To find the equation whose roots are smaller by 1 than those of the quadratic equation \(2x^2 - 5x + 6 = 0\), we can follow these steps:

### Step 1: Identify the coefficients of the given quadratic equation
The given equation is \(2x^2 - 5x + 6 = 0\). Here, we identify:
- \(a = 2\)
- \(b = -5\)
- \(c = 6\)

**Hint:** The coefficients \(a\), \(b\), and \(c\) are the numbers in front of \(x^2\), \(x\), and the constant term, respectively.

### Step 2: Calculate the sum and product of the roots
Let the roots of the equation be \(\alpha\) and \(\beta\). Using Vieta's formulas:
- The sum of the roots \(\alpha + \beta = -\frac{b}{a} = -\frac{-5}{2} = \frac{5}{2}\)
- The product of the roots \(\alpha \beta = \frac{c}{a} = \frac{6}{2} = 3\)

**Hint:** Vieta's formulas relate the coefficients of a polynomial to sums and products of its roots.

### Step 3: Determine the new roots
Since we want the new roots to be smaller by 1, we have:
- New roots: \(\alpha - 1\) and \(\beta - 1\)

### Step 4: Calculate the sum of the new roots
The sum of the new roots is:
\[ (\alpha - 1) + (\beta - 1) = \alpha + \beta - 2 = \frac{5}{2} - 2 = \frac{5}{2} - \frac{4}{2} = \frac{1}{2} \]

**Hint:** To find the new sum, simply subtract 2 from the original sum of the roots.

### Step 5: Calculate the product of the new roots
The product of the new roots is:
\[ (\alpha - 1)(\beta - 1) = \alpha \beta - (\alpha + \beta) + 1 = 3 - \frac{5}{2} + 1 \]
Calculating this gives:
\[ = 3 - \frac{5}{2} + \frac{2}{2} = 3 - \frac{3}{2} = \frac{6}{2} - \frac{3}{2} = \frac{3}{2} \]

**Hint:** Use the distributive property to expand the product and substitute the known values.

### Step 6: Form the new quadratic equation
The new quadratic equation can be formed using the sum and product of the new roots:
\[ x^2 - \text{(sum of roots)} \cdot x + \text{(product of roots)} = 0 \]
Substituting the values we found:
\[ x^2 - \frac{1}{2}x + \frac{3}{2} = 0 \]

### Step 7: Eliminate the fraction
To eliminate the fractions, multiply the entire equation by 2:
\[ 2x^2 - x + 3 = 0 \]

### Conclusion
Thus, the equation whose roots are smaller by 1 than those of \(2x^2 - 5x + 6 = 0\) is:
\[ \boxed{2x^2 - x + 3 = 0} \]