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### Q14. If [tex]\(\alpha, \beta\)[/tex] are zeroes of the polynomial [tex]\(p(x) = 5x^2 + 5x + 1\)[/tex], then find the values of:
(i) [tex]\(\alpha^2 + \beta^2\)[/tex]
(ii) [tex]\(\alpha^{-1} + \beta^{-1}\)[/tex]
(iii) [tex]\(\alpha^3 + \beta^3\)[/tex]
#### Solution:
Firstly, we recognize that [tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex] are the roots of the polynomial [tex]\(p(x) = 5x^2 + 5x + 1\)[/tex]. By Vieta's formulas for a quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex], the sum and product of the roots are given by:
- [tex]\(\alpha + \beta = -\frac{b}{a}\)[/tex]
- [tex]\(\alpha \beta = \frac{c}{a}\)[/tex]
For the given polynomial:
- [tex]\(a = 5\)[/tex]
- [tex]\(b = 5\)[/tex]
- [tex]\(c = 1\)[/tex]
Thus,
[tex]\[ \alpha + \beta = -\frac{5}{5} = -1 \][/tex]
[tex]\[ \alpha \beta = \frac{1}{5} \][/tex]
(i) Finding [tex]\(\alpha^2 + \beta^2\)[/tex]:
Using the identity [tex]\(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta\)[/tex]:
[tex]\[ \alpha^2 + \beta^2 = (-1)^2 - 2 \cdot \frac{1}{5} = 1 - \frac{2}{5} = \frac{5}{5} - \frac{2}{5} = \frac{3}{5} = 0.6 \][/tex]
(ii) Finding [tex]\(\alpha^{-1} + \beta^{-1}\)[/tex]:
To find [tex]\(\alpha^{-1} + \beta^{-1}\)[/tex], we use the relationship [tex]\(\alpha^{-1} + \beta^{-1} = \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta}\)[/tex]:
[tex]\[ \alpha^{-1} + \beta^{-1} = \frac{-1}{\frac{1}{5}} = -1 \cdot 5 = -5 \][/tex]
(iii) Finding [tex]\(\alpha^3 + \beta^3\)[/tex]:
Using the identity [tex]\(\alpha^3 + \beta^3 = (\alpha + \beta)((\alpha^2 + \beta^2) - \alpha \beta)\)[/tex]:
[tex]\[ \alpha^3 + \beta^3 = (-1)\left(0.6 - \frac{1}{5}\right) = (-1)\left(0.6 - 0.2\right) = (-1)(0.4) = -0.4 \][/tex]
Therefore, the values are:
- (i) [tex]\(\alpha^2 + \beta^2 = 0.6\)[/tex]
- (ii) [tex]\(\alpha^{-1} + \beta^{-1} = -5\)[/tex]
- (iii) [tex]\(\alpha^3 + \beta^3 = -0.4\)[/tex]
### Q15. The sum of the digits of a two-digit number is 9. The number obtained by reversing the order of digits of the given number exceeds the given number by 27. Find the given number.
#### Solution:
Let's denote the two-digit number as [tex]\(10a + b\)[/tex], where [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are the digits of the number.
Given conditions:
1. [tex]\(a + b = 9\)[/tex]
2. The number obtained by reversing the digits is [tex]\(10b + a\)[/tex], and this number exceeds the original number by 27.
Hence:
[tex]\[ (10b + a) - (10a + b) = 27 \implies 10b + a - 10a - b = 27 \implies 9b - 9a = 27 \implies b - a = 3 \][/tex]
We now have the system of equations:
1. [tex]\(a + b = 9\)[/tex]
2. [tex]\(b - a = 3\)[/tex]
To solve these equations simultaneously, add the two equations:
[tex]\[ (a + b) + (b - a) = 9 + 3 \implies 2b = 12 \implies b = 6 \][/tex]
Substitute [tex]\(b = 6\)[/tex] back into [tex]\(a + b = 9\)[/tex]:
[tex]\[ a + 6 = 9 \implies a = 3 \][/tex]
Thus, [tex]\(a = 3\)[/tex] and [tex]\(b = 6\)[/tex].
The original two-digit number is:
[tex]\[ 10a + b = 10(3) + 6 = 30 + 6 = 36 \][/tex]
Therefore, the given number is [tex]\(36\)[/tex].
### Summary:
1. For Q14:
- [tex]\(\alpha^2 + \beta^2 = 0.6\)[/tex]
- [tex]\(\alpha^{-1} + \beta^{-1} = -5\)[/tex]
- [tex]\(\alpha^3 + \beta^3 = -0.4\)[/tex]
2. For Q15:
- The given number is [tex]\(36\)[/tex].
### Q14. If [tex]\(\alpha, \beta\)[/tex] are zeroes of the polynomial [tex]\(p(x) = 5x^2 + 5x + 1\)[/tex], then find the values of:
(i) [tex]\(\alpha^2 + \beta^2\)[/tex]
(ii) [tex]\(\alpha^{-1} + \beta^{-1}\)[/tex]
(iii) [tex]\(\alpha^3 + \beta^3\)[/tex]
#### Solution:
Firstly, we recognize that [tex]\(\alpha\)[/tex] and [tex]\(\beta\)[/tex] are the roots of the polynomial [tex]\(p(x) = 5x^2 + 5x + 1\)[/tex]. By Vieta's formulas for a quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex], the sum and product of the roots are given by:
- [tex]\(\alpha + \beta = -\frac{b}{a}\)[/tex]
- [tex]\(\alpha \beta = \frac{c}{a}\)[/tex]
For the given polynomial:
- [tex]\(a = 5\)[/tex]
- [tex]\(b = 5\)[/tex]
- [tex]\(c = 1\)[/tex]
Thus,
[tex]\[ \alpha + \beta = -\frac{5}{5} = -1 \][/tex]
[tex]\[ \alpha \beta = \frac{1}{5} \][/tex]
(i) Finding [tex]\(\alpha^2 + \beta^2\)[/tex]:
Using the identity [tex]\(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta\)[/tex]:
[tex]\[ \alpha^2 + \beta^2 = (-1)^2 - 2 \cdot \frac{1}{5} = 1 - \frac{2}{5} = \frac{5}{5} - \frac{2}{5} = \frac{3}{5} = 0.6 \][/tex]
(ii) Finding [tex]\(\alpha^{-1} + \beta^{-1}\)[/tex]:
To find [tex]\(\alpha^{-1} + \beta^{-1}\)[/tex], we use the relationship [tex]\(\alpha^{-1} + \beta^{-1} = \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta}\)[/tex]:
[tex]\[ \alpha^{-1} + \beta^{-1} = \frac{-1}{\frac{1}{5}} = -1 \cdot 5 = -5 \][/tex]
(iii) Finding [tex]\(\alpha^3 + \beta^3\)[/tex]:
Using the identity [tex]\(\alpha^3 + \beta^3 = (\alpha + \beta)((\alpha^2 + \beta^2) - \alpha \beta)\)[/tex]:
[tex]\[ \alpha^3 + \beta^3 = (-1)\left(0.6 - \frac{1}{5}\right) = (-1)\left(0.6 - 0.2\right) = (-1)(0.4) = -0.4 \][/tex]
Therefore, the values are:
- (i) [tex]\(\alpha^2 + \beta^2 = 0.6\)[/tex]
- (ii) [tex]\(\alpha^{-1} + \beta^{-1} = -5\)[/tex]
- (iii) [tex]\(\alpha^3 + \beta^3 = -0.4\)[/tex]
### Q15. The sum of the digits of a two-digit number is 9. The number obtained by reversing the order of digits of the given number exceeds the given number by 27. Find the given number.
#### Solution:
Let's denote the two-digit number as [tex]\(10a + b\)[/tex], where [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are the digits of the number.
Given conditions:
1. [tex]\(a + b = 9\)[/tex]
2. The number obtained by reversing the digits is [tex]\(10b + a\)[/tex], and this number exceeds the original number by 27.
Hence:
[tex]\[ (10b + a) - (10a + b) = 27 \implies 10b + a - 10a - b = 27 \implies 9b - 9a = 27 \implies b - a = 3 \][/tex]
We now have the system of equations:
1. [tex]\(a + b = 9\)[/tex]
2. [tex]\(b - a = 3\)[/tex]
To solve these equations simultaneously, add the two equations:
[tex]\[ (a + b) + (b - a) = 9 + 3 \implies 2b = 12 \implies b = 6 \][/tex]
Substitute [tex]\(b = 6\)[/tex] back into [tex]\(a + b = 9\)[/tex]:
[tex]\[ a + 6 = 9 \implies a = 3 \][/tex]
Thus, [tex]\(a = 3\)[/tex] and [tex]\(b = 6\)[/tex].
The original two-digit number is:
[tex]\[ 10a + b = 10(3) + 6 = 30 + 6 = 36 \][/tex]
Therefore, the given number is [tex]\(36\)[/tex].
### Summary:
1. For Q14:
- [tex]\(\alpha^2 + \beta^2 = 0.6\)[/tex]
- [tex]\(\alpha^{-1} + \beta^{-1} = -5\)[/tex]
- [tex]\(\alpha^3 + \beta^3 = -0.4\)[/tex]
2. For Q15:
- The given number is [tex]\(36\)[/tex].
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