Answer :
Let’s solve the problem step-by-step:
### Part (a)
The quadratic equation is given as [tex]\(x^2 - px + q = 0\)[/tex] with roots [tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex].
#### (i) Write down the value of [tex]\(q\)[/tex]
Since [tex]\(\alpha\)[/tex] and \beta\ are the roots, we know from Vieta's formulas that:
- Sum of roots [tex]\( \alpha + \beta = p \)[/tex]
- Product of roots [tex]\( \alpha \beta = q \)[/tex]
We are given [tex]\( 2\alpha\beta = 3 \)[/tex], so:
[tex]\[ \alpha \beta = q = \frac{3}{2} \][/tex]
Thus, the value of [tex]\(q\)[/tex] is:
[tex]\[ q = \frac{3}{2} \][/tex]
#### (ii) Expression for [tex]\(p\)[/tex] in terms of [tex]\(k\)[/tex]
We are given:
[tex]\[ 4(\alpha^2 + \beta^2) = k^2 - 6k - 3 \][/tex]
First, recall the identity for the sum of squares of roots:
[tex]\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta \][/tex]
Substitute [tex]\( \alpha + \beta = p \)[/tex] and [tex]\( \alpha \beta = \frac{3}{2} \)[/tex]:
[tex]\[ \alpha^2 + \beta^2 = p^2 - 2 \left( \frac{3}{2} \right) \][/tex]
[tex]\[ \alpha^2 + \beta^2 = p^2 - 3 \][/tex]
Now, substitute this into the given equation:
[tex]\[ 4(p^2 - 3) = k^2 - 6k - 3 \][/tex]
[tex]\[ 4p^2 - 12 = k^2 - 6k - 3 \][/tex]
[tex]\[ 4p^2 = k^2 - 6k + 9 \][/tex]
[tex]\[ 4p^2 = (k - 3)^2 \][/tex]
Taking the positive value of the square root (since [tex]\( p > 0 \)[/tex]):
[tex]\[ 2p = k - 3 \][/tex]
[tex]\[ p = \frac{k - 3}{2} \][/tex]
Thus, the expression for [tex]\( p \)[/tex] in terms of [tex]\( k \)[/tex] is:
[tex]\[ p = \frac{k - 3}{2} \][/tex]
### Part (b)
Given the additional condition:
[tex]\[ 7 \alpha \beta = 3(\alpha + \beta) \][/tex]
We know [tex]\( \alpha \beta = \frac{3}{2} \)[/tex] and [tex]\( \alpha + \beta = p \)[/tex], so substitute these into the given condition:
[tex]\[ 7 \left( \frac{3}{2} \right) = 3p \][/tex]
[tex]\[ \frac{21}{2} = 3p \][/tex]
[tex]\[ p = \frac{21}{6} = \frac{7}{2} \][/tex]
We already have the expression for [tex]\( p \)[/tex] in terms of [tex]\( k \)[/tex]:
[tex]\[ \frac{k - 3}{2} = \frac{7}{2} \][/tex]
[tex]\[ k - 3 = 7 \][/tex]
[tex]\[ k = 10 \][/tex]
Thus, the value of [tex]\( k \)[/tex] is:
[tex]\[ k = 10 \][/tex]
### Part (c)
We need to form an equation with integer coefficients that has roots:
[tex]\[ \frac{\alpha}{\alpha + \beta} \text{ and } \frac{\beta}{\alpha + \beta} \][/tex]
Let’s denote [tex]\( \alpha + \beta = s \)[/tex]. Then the roots can be written as:
[tex]\[ \frac{\alpha}{s} \text{ and } \frac{\beta}{s} \][/tex]
In this case, these roots sum to:
[tex]\[ \frac{\alpha}{s} + \frac{\beta}{s} = \frac{\alpha + \beta}{s} = \frac{s}{s} = 1 \][/tex]
And the product of the roots is:
[tex]\[ \frac{\alpha}{s} \cdot \frac{\beta}{s} = \frac{\alpha \beta}{s^2} = \frac{\frac{3}{2}}{s^2} = \frac{3}{2s^2} \][/tex]
Since [tex]\( s = p = \frac{7}{2} \)[/tex], we substitute [tex]\( s \)[/tex]:
[tex]\[ \frac{3}{2s^2} = \frac{3}{2\left(\frac{7}{2}\right)^2} = \frac{3}{2 \cdot \frac{49}{4}} = \frac{3}{\frac{98}{4}} = \frac{3 \times 4}{98} = \frac{12}{98} = \frac{6}{49} \][/tex]
Thus, the roots sum to [tex]\(1\)[/tex] and have a product of [tex]\(\frac{6}{49}\)[/tex].
The quadratic equation with sum of roots 1 and product [tex]\(\frac{6}{49}\)[/tex] is:
[tex]\[ x^2 - x + \frac{6}{49} = 0 \][/tex]
To make the coefficients integers, multiply through by 49:
[tex]\[ 49x^2 - 49x + 6 = 0 \][/tex]
Thus, the final equation with integer coefficients is:
[tex]\[ 49x^2 - 49x + 6 = 0 \][/tex]
### Part (a)
The quadratic equation is given as [tex]\(x^2 - px + q = 0\)[/tex] with roots [tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex].
#### (i) Write down the value of [tex]\(q\)[/tex]
Since [tex]\(\alpha\)[/tex] and \beta\ are the roots, we know from Vieta's formulas that:
- Sum of roots [tex]\( \alpha + \beta = p \)[/tex]
- Product of roots [tex]\( \alpha \beta = q \)[/tex]
We are given [tex]\( 2\alpha\beta = 3 \)[/tex], so:
[tex]\[ \alpha \beta = q = \frac{3}{2} \][/tex]
Thus, the value of [tex]\(q\)[/tex] is:
[tex]\[ q = \frac{3}{2} \][/tex]
#### (ii) Expression for [tex]\(p\)[/tex] in terms of [tex]\(k\)[/tex]
We are given:
[tex]\[ 4(\alpha^2 + \beta^2) = k^2 - 6k - 3 \][/tex]
First, recall the identity for the sum of squares of roots:
[tex]\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta \][/tex]
Substitute [tex]\( \alpha + \beta = p \)[/tex] and [tex]\( \alpha \beta = \frac{3}{2} \)[/tex]:
[tex]\[ \alpha^2 + \beta^2 = p^2 - 2 \left( \frac{3}{2} \right) \][/tex]
[tex]\[ \alpha^2 + \beta^2 = p^2 - 3 \][/tex]
Now, substitute this into the given equation:
[tex]\[ 4(p^2 - 3) = k^2 - 6k - 3 \][/tex]
[tex]\[ 4p^2 - 12 = k^2 - 6k - 3 \][/tex]
[tex]\[ 4p^2 = k^2 - 6k + 9 \][/tex]
[tex]\[ 4p^2 = (k - 3)^2 \][/tex]
Taking the positive value of the square root (since [tex]\( p > 0 \)[/tex]):
[tex]\[ 2p = k - 3 \][/tex]
[tex]\[ p = \frac{k - 3}{2} \][/tex]
Thus, the expression for [tex]\( p \)[/tex] in terms of [tex]\( k \)[/tex] is:
[tex]\[ p = \frac{k - 3}{2} \][/tex]
### Part (b)
Given the additional condition:
[tex]\[ 7 \alpha \beta = 3(\alpha + \beta) \][/tex]
We know [tex]\( \alpha \beta = \frac{3}{2} \)[/tex] and [tex]\( \alpha + \beta = p \)[/tex], so substitute these into the given condition:
[tex]\[ 7 \left( \frac{3}{2} \right) = 3p \][/tex]
[tex]\[ \frac{21}{2} = 3p \][/tex]
[tex]\[ p = \frac{21}{6} = \frac{7}{2} \][/tex]
We already have the expression for [tex]\( p \)[/tex] in terms of [tex]\( k \)[/tex]:
[tex]\[ \frac{k - 3}{2} = \frac{7}{2} \][/tex]
[tex]\[ k - 3 = 7 \][/tex]
[tex]\[ k = 10 \][/tex]
Thus, the value of [tex]\( k \)[/tex] is:
[tex]\[ k = 10 \][/tex]
### Part (c)
We need to form an equation with integer coefficients that has roots:
[tex]\[ \frac{\alpha}{\alpha + \beta} \text{ and } \frac{\beta}{\alpha + \beta} \][/tex]
Let’s denote [tex]\( \alpha + \beta = s \)[/tex]. Then the roots can be written as:
[tex]\[ \frac{\alpha}{s} \text{ and } \frac{\beta}{s} \][/tex]
In this case, these roots sum to:
[tex]\[ \frac{\alpha}{s} + \frac{\beta}{s} = \frac{\alpha + \beta}{s} = \frac{s}{s} = 1 \][/tex]
And the product of the roots is:
[tex]\[ \frac{\alpha}{s} \cdot \frac{\beta}{s} = \frac{\alpha \beta}{s^2} = \frac{\frac{3}{2}}{s^2} = \frac{3}{2s^2} \][/tex]
Since [tex]\( s = p = \frac{7}{2} \)[/tex], we substitute [tex]\( s \)[/tex]:
[tex]\[ \frac{3}{2s^2} = \frac{3}{2\left(\frac{7}{2}\right)^2} = \frac{3}{2 \cdot \frac{49}{4}} = \frac{3}{\frac{98}{4}} = \frac{3 \times 4}{98} = \frac{12}{98} = \frac{6}{49} \][/tex]
Thus, the roots sum to [tex]\(1\)[/tex] and have a product of [tex]\(\frac{6}{49}\)[/tex].
The quadratic equation with sum of roots 1 and product [tex]\(\frac{6}{49}\)[/tex] is:
[tex]\[ x^2 - x + \frac{6}{49} = 0 \][/tex]
To make the coefficients integers, multiply through by 49:
[tex]\[ 49x^2 - 49x + 6 = 0 \][/tex]
Thus, the final equation with integer coefficients is:
[tex]\[ 49x^2 - 49x + 6 = 0 \][/tex]
