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Sagot :
Let's solve the given problem step-by-step.
## Part (a)
### (i) Write down the value of [tex]\( q \)[/tex].
We know that for a quadratic equation [tex]\( x^2 - p x + q = 0 \)[/tex] with roots [tex]\( \alpha \)[/tex] and [tex]\( \beta \)[/tex]:
- [tex]\( \alpha + \beta = p \)[/tex]
- [tex]\( \alpha \beta = q \)[/tex]
Given [tex]\( 2 \alpha \beta = 3 \)[/tex], we can find [tex]\( \alpha \beta \)[/tex]:
[tex]\[ 2 \alpha \beta = 3 \implies \alpha \beta = \frac{3}{2} \][/tex]
Thus, the value of [tex]\( q \)[/tex] is:
[tex]\[ q = \frac{3}{2} \][/tex]
### (ii) Find an expression, in terms of [tex]\( k \)[/tex], for [tex]\( p \)[/tex].
We are also given the expression:
[tex]\[ 4 (\alpha^2 + \beta^2) = k^2 - 6k - 3 \][/tex]
We use the identity [tex]\( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta \)[/tex]:
[tex]\[ 4 (\alpha^2 + \beta^2) = 4 \left( (\alpha + \beta)^2 - 2 \alpha \beta \right) \][/tex]
[tex]\[ 4 \left( (\alpha + \beta)^2 - 2 \alpha \beta \right) = k^2 - 6k - 3 \][/tex]
We already know that [tex]\( \alpha + \beta = p \)[/tex] and [tex]\( \alpha \beta = \frac{3}{2} \)[/tex]:
[tex]\[ 4 \left( p^2 - 2 \times \frac{3}{2} \right) = k^2 - 6k - 3 \][/tex]
[tex]\[ 4 \left( p^2 - 3 \right) = k^2 - 6k - 3 \][/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 square root of both sides, we get:
[tex]\[ 2p = k - 3 \][/tex]
Thus:
[tex]\[ p = \frac{k - 3}{2} \][/tex]
## Part (b)
Given also that [tex]\( 7 \alpha \beta = 3 (\alpha + \beta) \)[/tex]:
[tex]\[ 7 \times \frac{3}{2} = 3 (\alpha + \beta) \][/tex]
[tex]\[ \frac{21}{2} = 3p \][/tex]
[tex]\[ p = \frac{21}{6} \][/tex]
[tex]\[ p = \frac{7}{2} \][/tex]
From part (a), we already have [tex]\( p = \frac{k - 3}{2} \)[/tex]:
[tex]\[ \frac{k - 3}{2} = \frac{7}{2} \][/tex]
[tex]\[ k - 3 = 7 \][/tex]
[tex]\[ k = 10 \][/tex]
## Part (c)
To form an equation with integer coefficients that has roots [tex]\( \frac{\alpha}{\alpha + \beta} \)[/tex] and [tex]\( \frac{\beta}{\alpha + \beta} \)[/tex], we start by recognizing the roots of the new polynomial. Let [tex]\( \alpha + \beta = p \)[/tex]:
[tex]\[ p = \frac{10 - 3}{2} = \frac{7}{2} \][/tex]
Since [tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex] are the roots of the original polynomial, the sum [tex]\(\frac{\alpha}{\alpha + \beta} + \frac{\beta}{\alpha + \beta} = 1\)[/tex].
Let the roots be [tex]\( \frac{\alpha}{\alpha + \beta} \)[/tex] and [tex]\( \frac{\beta}{\alpha + \beta} \)[/tex]:
[tex]\[ \frac{\alpha}{p} + \frac{\beta}{p} = 1 \][/tex]
The polynomial we seek is:
[tex]\[ y^2 - y + r = 0 \][/tex]
Given that the product of roots [tex]\( \frac{\alpha}{\alpha + \beta} \times \frac{\beta}{\alpha + \beta} \)[/tex] is equivalent to:
[tex]\[ \frac{\alpha \beta}{(\alpha + \beta)^2} \][/tex]
[tex]\[ \frac{\frac{3}{2}}{\left(\frac{7}{2}\right)^2} \][/tex]
[tex]\[ \frac{\frac{3}{2}}{\frac{49}{4}} \][/tex]
[tex]\[ \frac{3}{2} \times \frac{4}{49} \][/tex]
[tex]\[ \frac{12}{98} = \frac{6}{49} \][/tex]
Thus, [tex]\( r = \frac{6}{49} \)[/tex], but since we want integer coefficients, we can multiply through by 49 to clear the denominator:
[tex]\[ 49y^2 - 49y + 6 = 0 \][/tex]
Therefore, the polynomial equation with integer coefficients is:
[tex]\[ 49y^2 - 49y + 6 = 0 \][/tex]
## Part (a)
### (i) Write down the value of [tex]\( q \)[/tex].
We know that for a quadratic equation [tex]\( x^2 - p x + q = 0 \)[/tex] with roots [tex]\( \alpha \)[/tex] and [tex]\( \beta \)[/tex]:
- [tex]\( \alpha + \beta = p \)[/tex]
- [tex]\( \alpha \beta = q \)[/tex]
Given [tex]\( 2 \alpha \beta = 3 \)[/tex], we can find [tex]\( \alpha \beta \)[/tex]:
[tex]\[ 2 \alpha \beta = 3 \implies \alpha \beta = \frac{3}{2} \][/tex]
Thus, the value of [tex]\( q \)[/tex] is:
[tex]\[ q = \frac{3}{2} \][/tex]
### (ii) Find an expression, in terms of [tex]\( k \)[/tex], for [tex]\( p \)[/tex].
We are also given the expression:
[tex]\[ 4 (\alpha^2 + \beta^2) = k^2 - 6k - 3 \][/tex]
We use the identity [tex]\( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta \)[/tex]:
[tex]\[ 4 (\alpha^2 + \beta^2) = 4 \left( (\alpha + \beta)^2 - 2 \alpha \beta \right) \][/tex]
[tex]\[ 4 \left( (\alpha + \beta)^2 - 2 \alpha \beta \right) = k^2 - 6k - 3 \][/tex]
We already know that [tex]\( \alpha + \beta = p \)[/tex] and [tex]\( \alpha \beta = \frac{3}{2} \)[/tex]:
[tex]\[ 4 \left( p^2 - 2 \times \frac{3}{2} \right) = k^2 - 6k - 3 \][/tex]
[tex]\[ 4 \left( p^2 - 3 \right) = k^2 - 6k - 3 \][/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 square root of both sides, we get:
[tex]\[ 2p = k - 3 \][/tex]
Thus:
[tex]\[ p = \frac{k - 3}{2} \][/tex]
## Part (b)
Given also that [tex]\( 7 \alpha \beta = 3 (\alpha + \beta) \)[/tex]:
[tex]\[ 7 \times \frac{3}{2} = 3 (\alpha + \beta) \][/tex]
[tex]\[ \frac{21}{2} = 3p \][/tex]
[tex]\[ p = \frac{21}{6} \][/tex]
[tex]\[ p = \frac{7}{2} \][/tex]
From part (a), we already have [tex]\( p = \frac{k - 3}{2} \)[/tex]:
[tex]\[ \frac{k - 3}{2} = \frac{7}{2} \][/tex]
[tex]\[ k - 3 = 7 \][/tex]
[tex]\[ k = 10 \][/tex]
## Part (c)
To form an equation with integer coefficients that has roots [tex]\( \frac{\alpha}{\alpha + \beta} \)[/tex] and [tex]\( \frac{\beta}{\alpha + \beta} \)[/tex], we start by recognizing the roots of the new polynomial. Let [tex]\( \alpha + \beta = p \)[/tex]:
[tex]\[ p = \frac{10 - 3}{2} = \frac{7}{2} \][/tex]
Since [tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex] are the roots of the original polynomial, the sum [tex]\(\frac{\alpha}{\alpha + \beta} + \frac{\beta}{\alpha + \beta} = 1\)[/tex].
Let the roots be [tex]\( \frac{\alpha}{\alpha + \beta} \)[/tex] and [tex]\( \frac{\beta}{\alpha + \beta} \)[/tex]:
[tex]\[ \frac{\alpha}{p} + \frac{\beta}{p} = 1 \][/tex]
The polynomial we seek is:
[tex]\[ y^2 - y + r = 0 \][/tex]
Given that the product of roots [tex]\( \frac{\alpha}{\alpha + \beta} \times \frac{\beta}{\alpha + \beta} \)[/tex] is equivalent to:
[tex]\[ \frac{\alpha \beta}{(\alpha + \beta)^2} \][/tex]
[tex]\[ \frac{\frac{3}{2}}{\left(\frac{7}{2}\right)^2} \][/tex]
[tex]\[ \frac{\frac{3}{2}}{\frac{49}{4}} \][/tex]
[tex]\[ \frac{3}{2} \times \frac{4}{49} \][/tex]
[tex]\[ \frac{12}{98} = \frac{6}{49} \][/tex]
Thus, [tex]\( r = \frac{6}{49} \)[/tex], but since we want integer coefficients, we can multiply through by 49 to clear the denominator:
[tex]\[ 49y^2 - 49y + 6 = 0 \][/tex]
Therefore, the polynomial equation with integer coefficients is:
[tex]\[ 49y^2 - 49y + 6 = 0 \][/tex]
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