If `x^2+p x+q=0` is the quadratic equation whose roots are `a-2a n d b-2` where `aa n db` are the roots of `x^2-3x+1=0,` then `p-1,q=5` b. `p=1,1=-5` c. `p=-1,q=1` d. `p=1,q=-1`

Options

A. `p = 1, q = 5`
B. `p = 1, q = -5`
C. `p = -1, q = 1`
D. `p = 1, q = -1`

Answer

To solve the problem step by step, we will start by identifying the roots of the given quadratic equation and then relate them to the coefficients \( p \) and \( q \).

### Step 1: Find the roots of the equation \( x^2 - 3x + 1 = 0 \)

To find the roots \( a \) and \( b \), we can use the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a = 1 \), \( b = -3 \), and \( c = 1 \).

Calculating the discriminant:

\[ b^2 - 4ac = (-3)^2 - 4 \cdot 1 \cdot 1 = 9 - 4 = 5 \]

Now, substituting into the quadratic formula:

\[ x = \frac{3 \pm \sqrt{5}}{2} \]

Thus, the roots are:

\[ a = \frac{3 + \sqrt{5}}{2}, \quad b = \frac{3 - \sqrt{5}}{2} \]

### Step 2: Determine the new roots \( a - 2 \) and \( b - 2 \)

Now, we find the new roots:

\[ \text{New roots: } a - 2 = \frac{3 + \sqrt{5}}{2} - 2 = \frac{-1 + \sqrt{5}}{2} \]
\[ b - 2 = \frac{3 - \sqrt{5}}{2} - 2 = \frac{-1 - \sqrt{5}}{2} \]

### Step 3: Calculate the sum and product of the new roots

The sum of the new roots \( (a - 2) + (b - 2) \):

\[ = \left(\frac{-1 + \sqrt{5}}{2}\right) + \left(\frac{-1 - \sqrt{5}}{2}\right) = \frac{-2}{2} = -1 \]

The product of the new roots \( (a - 2)(b - 2) \):

\[ = \left(\frac{-1 + \sqrt{5}}{2}\right) \left(\frac{-1 - \sqrt{5}}{2}\right) = \frac{(-1)^2 - (\sqrt{5})^2}{4} = \frac{1 - 5}{4} = \frac{-4}{4} = -1 \]

### Step 4: Relate the sum and product to \( p \) and \( q \)

From Vieta's formulas, for the quadratic equation \( x^2 + px + q = 0 \):

- The sum of the roots \( -(p) = -1 \) implies \( p = 1 \).
- The product of the roots \( q = -1 \).

### Conclusion

Thus, we have determined:

\[ p = 1, \quad q = -1 \]

The correct answer is option **d. \( p = 1, q = -1 \)**.

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