Quadratic equation lecture 3
Quadratic equation lecture 3
Question
Let α , β \alpha, \beta α , β x 2 x^2 x 2 α / 2 , 2 β \alpha / 2,2 \beta α /2 , 2 β x 2 x^2 x 2
(1) 2 9 ( p − q ) ( 2 q − p ) \frac{2}{9}(p-q)(2 q-p) 9 2 ( p − q ) ( 2 q − p )
(2) 2 9 ( q − p ) ( 2 p − q ) \frac{2}{9}(q-p)(2 p-q) 9 2 ( q − p ) ( 2 p − q )
(3) 2 9 ( q − 2 p ) ( 2 q − p ) \frac{2}{9}(q-2p)(2 q-p) 9 2 ( q − 2 p ) ( 2 q − p )
(4) 2 9 ( 2 p − q ) ( 2 q − p ) \frac{2}{9}(2 p-q)(2 q-p) 9 2 ( 2 p − q ) ( 2 q − p )
Quadratic equation lecture 3
Solution
α + β = p \alpha + \beta = p α + β = p
α β = r \alpha \beta = r α β = r
α 2 + 2 β = q \frac{\alpha}{2} + 2 \beta = q 2 α + 2 β = q
α β = r \alpha \beta = r α β = r
α + 4 β = 2 q \alpha + 4 \beta = 2 q α + 4 β = 2 q 3 β = 2 q − p 3 \beta=2 q-p 3 β = 2 q − p
Quadratic equation lecture 3
Solution
⇒ β = 2 q − p 3 \Rightarrow \beta=\frac{2 q-p}{3} ⇒ β = 3 2 q − p
α = p − β = p − 2 q − p 3 \alpha=p-\beta =p-\frac{2 q-p}{3} α = p − β = p − 3 2 q − p
= 2 ( 2 p − q ) 3 =\frac{2(2 p-q)}{3} = 3 2 ( 2 p − q )
α β = r \alpha \beta =r α β = r
∴ r = 2 3 ( 2 p − q ) ( 2 q − p ) 3 \therefore r =\frac{2}{3}(2 p-q) \frac{(2 q-p)}{3} ∴ r = 3 2 ( 2 p − q ) 3 ( 2 q − p )
= 2 9 ( 2 p − q ) ( 2 q − p ) =\frac{2}{9}(2 p-q)(2 q-p) = 9 2 ( 2 p − q ) ( 2 q − p )
Quadratic equation lecture 3
Question
Let α , β \alpha, \beta α , β x 2 − 5 x + 3 = 0 x^2-5x+3=0 x 2 − 5 x + 3 = 0 α / β \alpha/\beta α / β β / α \beta/\alpha β / α
(1) 3 x 2 − 19 x + 3 = 0 3x^2-19x+3=0 3 x 2 − 19 x + 3 = 0
(2) 3 x 2 + 19 x + 3 = 0 3x^2+19x+3=0 3 x 2 + 19 x + 3 = 0
(3) 3 x 2 + 5 x + 3 = 0 3x^2+5x+3=0 3 x 2 + 5 x + 3 = 0
(4) 3 x 2 − 5 x + 3 = 0 3x^2-5x+3=0 3 x 2 − 5 x + 3 = 0
Quadratic equation lecture 3
Solution
α + β = 5 \alpha + \beta = 5 α + β = 5
α β = 3 \alpha \beta = 3 α β = 3
α β + β α = α 2 + β 2 α β \frac{\alpha}{\beta}+\frac{\beta}{\alpha} =\frac{\alpha^2+\beta^2}{\alpha \beta} β α + α β = α β α 2 + β 2 = ( α + β ) 2 − 2 α β α β = 25 − 6 3 = 19 3 =\frac{(\alpha+\beta)^2-2 \alpha \beta}{\alpha \beta} ~ =\frac{25-6}{3}=\frac{19}{3} = α β ( α + β ) 2 − 2 α β = 3 25 − 6 = 3 19
α β β α = 1 \frac{\alpha}{\beta} \frac{\beta}{\alpha}=1 β α α β = 1
x 2 − 19 3 x + 1 = 0 x^2-\frac{19}{3}x +1=0 x 2 − 3 19 x + 1 = 0
3 x 2 − 19 x + 3 = 0 3 x^2-19 x +3=0 3 x 2 − 19 x + 3 = 0
Quadratic equation lecture 3
Question
Let p, q be two real numbers such that p ≠ 0 , p 3 ≠ ± q p \neq 0, p^3 \neq \pm q p = 0 , p 3 = ± q α , β \alpha, \beta α , β α + β = − p \alpha+\beta=-p α + β = − p α 3 + β 3 = q \alpha^3+\beta^3 = q α 3 + β 3 = q α / β \alpha / \beta α / β β / α \beta / \alpha β / α
(1) ( p 3 + q ) x 2 − ( p 3 + 2 q ) x + ( p 3 + q ) = 0 (p^3+q)x^2-(p^3+2q)x+(p^3+q) = 0 ( p 3 + q ) x 2 − ( p 3 + 2 q ) x + ( p 3 + q ) = 0
(2) ( p 3 + q ) x 2 − ( p 3 − 2 q ) x + ( p 3 + q ) = 0 (p^3+q)x^2-(p^3-2q)x+(p^3+q) = 0 ( p 3 + q ) x 2 − ( p 3 − 2 q ) x + ( p 3 + q ) = 0
(3) ( p 3 − q ) x 2 − ( 5 p 3 − 2 q ) x + ( p 3 − q ) = 0 (p^3-q)x^2-(5p^3-2q)x+(p^3-q) = 0 ( p 3 − q ) x 2 − ( 5 p 3 − 2 q ) x + ( p 3 − q ) = 0
(4) ( p 3 − q ) x 2 − ( 5 p 3 + 2 q ) x + ( p 3 − q ) = 0 (p^3-q)x^2-(5p^3+2q)x+(p^3-q) = 0 ( p 3 − q ) x 2 − ( 5 p 3 + 2 q ) x + ( p 3 − q ) = 0
Quadratic equation lecture 3
Solution
x 2 − ( α β + β α ) x + 1 = 0 x^2-(\frac{\alpha}{\beta}+\frac{\beta}{\alpha})x+1=0 x 2 − ( β α + α β ) x + 1 = 0
α β + β α = α 2 + β 2 α β = ( α + β ) 2 − 2 α β α β \frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha \beta}=\frac{(\alpha+\beta)^2-2 \alpha \beta}{\alpha \beta} β α + α β = α β α 2 + β 2 = α β ( α + β ) 2 − 2 α β
( α + β ) 3 = α 3 + β 3 + 3 α β ( α + β ) (\alpha+\beta)^3 =\alpha^3+\beta^3+3 \alpha \beta(\alpha+\beta) ( α + β ) 3 = α 3 + β 3 + 3 α β ( α + β )
− p 3 = q − 3 α β p -p^3 =q-3 \alpha \beta p − p 3 = q − 3 α βp
⇒ α β = p 3 + q 3 p \Rightarrow \alpha \beta =\frac{p^3+q}{3 p} ⇒ α β = 3 p p 3 + q
Quadratic equation lecture 3
Solution
α β + β α = p 2 − 2 p 3 + q 3 p p 3 + q 3 p = 3 p 3 − 2 p 3 − 2 q ( p 3 + q ) \frac{\alpha}{\beta}+\frac{\beta}{\alpha} =\frac{p^2-2 \frac{p^3+q}{3 p}}{\frac{p^3+q}{3 p}}=\frac{3 p^3-2 p^3-2 q}{(p^3+q)} β α + α β = 3 p p 3 + q p 2 − 2 3 p p 3 + q = ( p 3 + q ) 3 p 3 − 2 p 3 − 2 q
= p 3 − 2 q p 3 + q =\frac{p^3-2 q}{p^3+q} = p 3 + q p 3 − 2 q
x 2 − ( p 3 − 2 q p 3 + q ) x + 1 = 0 x^2-(\frac{p^3-2 q}{p^3+q})x+1=0 x 2 − ( p 3 + q p 3 − 2 q ) x + 1 = 0
( p 3 + q ) x 2 − ( p 3 − 2 q ) x + p 3 + q = 0 (p^3+q)x^2-(p^3-2 q)x+p^3+q=0 ( p 3 + q ) x 2 − ( p 3 − 2 q ) x + p 3 + q = 0
Quadratic equation lecture 3
Question
Let α , β \alpha, \beta α , β x 2 − 6 x − 2 = 0 x^2-6x-2=0 x 2 − 6 x − 2 = 0 α > β \alpha>\beta α > β a n = α n − β n a_n=\alpha^n - \beta^n a n = α n − β n n ≥ 1 {n \geq 1} n ≥ 1 a 10 − 2 a 8 2 a 9 \frac{a_{10}-2a_{8}}{2a_{9}} 2 a 9 a 10 − 2 a 8
(1) 1
(2) 2
(3) 3
(4) 4
Quadratic equation lecture 3
Solution
α 2 − 6 α − 2 = 0 [ α 2 − 2 = 6 α ] \alpha^2-6 \alpha-2=0 \quad \biggl[\alpha^2-2=6 \alpha\biggl] α 2 − 6 α − 2 = 0 [ α 2 − 2 = 6 α ]
β 2 − 6 β − 2 = 0 [ β 2 − 2 = 6 β ] \beta^2-6 \beta-2=0 \quad \biggl[\beta^2-2=6 \beta\biggl] β 2 − 6 β − 2 = 0 [ β 2 − 2 = 6 β ]
a 10 − 2 a 8 2 a 9 = α 10 − β 10 − 2 α 8 + 2 β 8 2 α 9 − 2 β 9 \frac{a_{10}-2a_8}{2a_9} =\frac{\alpha^{10}-\beta^{10}-2 \alpha^8+2 \beta^8}{2 \alpha^9-2 \beta^9} 2 a 9 a 10 − 2 a 8 = 2 α 9 − 2 β 9 α 10 − β 10 − 2 α 8 + 2 β 8
= α 8 ( α 2 − 2 ) − β 8 ( β 2 − 2 ) 2 ( α 9 − β 9 ) =\frac{\alpha^8(\alpha^2-2)-\beta^8(\beta^2-2)}{2(\alpha^9-\beta^9)} = 2 ( α 9 − β 9 ) α 8 ( α 2 − 2 ) − β 8 ( β 2 − 2 )
= 6 α 9 − 6 β 9 2 ( α 9 − β 9 ) = 6 2 = 3 =\frac{6 \alpha^9-6 \beta^9}{2(\alpha^9-\beta^9)}=\frac{6}{2}=3 = 2 ( α 9 − β 9 ) 6 α 9 − 6 β 9 = 2 6 = 3
Quadratic equation lecture 3
Question
Paragraph : Let p, q be integers and α , β \alpha, \beta α , β x 2 − x − 1 = 0 x^2-x-1 = 0 x 2 − x − 1 = 0 α ≠ β \alpha \neq \beta α = β a n = p α n + q β n a_{n} = {p \alpha}^n+q \beta^n a n = p α n + q β n n ≠ 0 n\neq0 n = 0
Fact : If a, b are rational numbers such that a + b 5 = 0 a+b\sqrt{5}=0 a + b 5 = 0
Question A : The value of a 12 a_{12} a 12
(1) a 11 − a 10 a_{11}-a_{10} a 11 − a 10 a 11 + a 10 a_{11} + a_{10} a 11 + a 10 2 a 11 + a 10 2a_{11} + a_{10} 2 a 11 + a 10 a 11 + 2 a 10 a_{11} + 2a_{10} a 11 + 2 a 10
Quadratic equation lecture 3
Question
Question B : If a 4 a_4 a 4
(1) 21
(2) 14
(3) 07
(4) 12
Quadratic equation lecture 3
Solution
A) a 12 = p α 12 + q β 12 ~ a_{12}=p \alpha^{12}+q \beta^{12} a 12 = p α 12 + q β 12
= p α 10 ⋅ α 2 + q β 10 ⋅ β 2 =p\alpha^{10} \cdot \alpha^2+q \beta^{10} \cdot \beta^2 = p α 10 ⋅ α 2 + q β 10 ⋅ β 2
= p α 10 ( α + 1 ) + q β 10 ( β + 1 ) =p\alpha^{10}(\alpha+1)+q \beta^{10}(\beta+1) = p α 10 ( α + 1 ) + q β 10 ( β + 1 )
= p α 11 + p α 10 + q β 11 + q β 10 =p\alpha^{11}+p \alpha^{10}+q \beta^{11}+q \beta^{10} = p α 11 + p α 10 + q β 11 + q β 10
Option (2) is correct.
[ α 2 − α − 1 = 0 ⇒ α 2 = α + 1 β 2 − β − 1 = 0 ⇒ β 2 = β + 1 ] \begin{bmatrix} ~ \alpha^2-\alpha-1=0 ~ \\\
~ \Rightarrow \alpha^2=\alpha+1 ~ \\\\
~ \beta^2-\beta-1=0\\\
~ \Rightarrow \beta^2=\beta+1 ~ \end{bmatrix} α 2 − α − 1 = 0 ⇒ α 2 = α + 1 β 2 − β − 1 = 0 ⇒ β 2 = β + 1
Quadratic equation lecture 3
Solution
B) a 4 = 28 ~ a_4=28 a 4 = 28
a 4 = a 3 + a 2 a_4=a_3+a_2 a 4 = a 3 + a 2
= a 2 + a 1 + a 1 + a 0 =a_2+a_1+a_1+a_0 = a 2 + a 1 + a 1 + a 0
= a 1 + a 0 + 2 a 1 + a 0 =a_1+a_0+2a_1+a_0 = a 1 + a 0 + 2 a 1 + a 0
= 3 a 1 + 2 a 0 =3a_1+2a_0 = 3 a 1 + 2 a 0
a 4 = 3 p α + 3 q β + 2 p a_4=3p \alpha+3 q \beta+2 p a 4 = 3 p α + 3 qβ + 2 p + + +
[ a 4 = p α 4 + q β 4 = p α 2 ( α + 1 ) + q β 2 ( β + 1 ) = ( p α 3 + p α 2 + 2 β 3 + q β 2 = a 3 + a 2 a n = a n − 1 + a n − 2 ∀ n ⩾ 2 a 0 = p + q a 1 = p α + q β ] \left[\begin{array}{l}a_4=p \alpha^4+q \beta^4=p \alpha^2(\alpha+1)+q \beta^2(\beta+1) \\\ =\left(p \alpha^3+p \alpha^2+2 \beta^3+q \beta^2\right. =a_3+a_2 \\\ a_n=a_{n-1}+a_{n-2} \\\ \forall n \geqslant 2 \\\ a_0=p+q \\\ a_1=p \alpha+q \beta\end{array}\right] a 4 = p α 4 + q β 4 = p α 2 ( α + 1 ) + q β 2 ( β + 1 ) = ( p α 3 + p α 2 + 2 β 3 + q β 2 = a 3 + a 2 a n = a n − 1 + a n − 2 ∀ n ⩾ 2 a 0 = p + q a 1 = p α + qβ
Quadratic equation lecture 3
Solution
x 2 − x − 1 = 0 x^2-x-1=0 x 2 − x − 1 = 0
1 ± 5 2 \frac{1 \pm \sqrt{5}}{2} 2 1 ± 5
α = 1 + 5 2 \alpha=\frac{1+\sqrt{5}}{2} α = 2 1 + 5
β = 1 − 5 2 \beta=\frac{1-\sqrt{5}}{2} β = 2 1 − 5
3 p α + 3 q β + 2 p + 2 q = 28 3p\alpha+3q\beta+2p+2q=28 3 p α + 3 qβ + 2 p + 2 q = 28
Quadratic equation lecture 3
Solution
3 p ( 1 + 5 2 ) + 3 q ( 1 − 5 2 ) + 2 p + 2 q = 28 3p(\frac{1+\sqrt{5}}{2})+3q(\frac{1-\sqrt{5}}{2})+2p+2q=28 3 p ( 2 1 + 5 ) + 3 q ( 2 1 − 5 ) + 2 p + 2 q = 28
⇒ ( 3 p + 3 p 5 + 3 q − 3 q 5 + 4 p + 4 q = 56 \Rightarrow(3p+3p \sqrt{5}+3q-3q \sqrt{5}+4p+4q=56 ⇒ ( 3 p + 3 p 5 + 3 q − 3 q 5 + 4 p + 4 q = 56
3p - 3q = 0
⇒ \Rightarrow ⇒
7p + 7q = 56
⇒ \Rightarrow ⇒
∴ \therefore ∴
∴ \therefore ~ ∴
= 4 + 8= 12
Quadratic equation lecture 3
Question
Let α , β \alpha, \beta α , β x 2 − x − 1 = 0 x^2-x-1=0 x 2 − x − 1 = 0 α > β \alpha>\beta α > β a n = α n − β n α − β a_n=\frac{\alpha^n-\beta^n}{\alpha-\beta} a n = α − β α n − β n n ≥ 1 , b 1 = 1 n \geq 1, b_1=1 n ≥ 1 , b 1 = 1 b n = a n − 1 + a n + 1 b_n={a_{n-1}+a_{n+1}} b n = a n − 1 + a n + 1 n ≥ 2 n \geq 2 n ≥ 2
(1) a 1 + a 2 + ⋯ + a n = a n + 2 − 1 a_1+a_2+\cdots+a_n=a_{n+2}-1 a 1 + a 2 + ⋯ + a n = a n + 2 − 1 n ≥ 1 n \geq 1 n ≥ 1
(2) b n = α n + β n b_n=\alpha^n+\beta^n b n = α n + β n n ≥ 1 n \geq 1 n ≥ 1
(3) ∑ n ≥ 1 b n 10 n = 8 89 \sum_{n\geq1} \frac{b_n}{10^n}=\frac{8}{89} ∑ n ≥ 1 1 0 n b n = 89 8
(4) ∑ n ≥ 1 a n 10 n = 10 89 \sum_{n\geq1} \frac{a_n}{10^n}=\frac{10}{89} ∑ n ≥ 1 1 0 n a n = 89 10
Quadratic equation lecture 3
Solution
1 ± 5 2 \frac{1\pm\sqrt{5}}{2} 2 1 ± 5
α = 1 + 5 2 \alpha=\frac{1+\sqrt{5}}{2} α = 2 1 + 5
β = 1 − 5 2 \beta=\frac{1-\sqrt{5}}{2} β = 2 1 − 5
b n = a n − 1 + a n + 1 ∀ n ⩾ 2 b_n=a_{n-1}+a_{n+1} \forall n \geqslant 2 b n = a n − 1 + a n + 1 ∀ n ⩾ 2
= α n − 1 − β n − 1 α − β + α n + 1 − β n + 1 α − β =\frac{\alpha^{n-1}-\beta^{n-1}}{\alpha-\beta}+\frac{\alpha^{n+1}-\beta^{n+1}}{\alpha-\beta} = α − β α n − 1 − β n − 1 + α − β α n + 1 − β n + 1 = α n − 1 ( α 2 + 1 ) − β n − 1 ( β 2 + 1 ) α − β =\frac{\alpha^{n-1}(\alpha^2+1)-\beta^{n-1}(\beta^2+1)}{\alpha-\beta} = α − β α n − 1 ( α 2 + 1 ) − β n − 1 ( β 2 + 1 )
= α n − 1 ( 5 α ) + β n − 1 ( 5 β ) 5 =\frac{\alpha^{n-1}(\sqrt{5} \alpha)+\beta^{n-1}(\sqrt{5} \beta)}{\sqrt{5}} = 5 α n − 1 ( 5 α ) + β n − 1 ( 5 β ) = α n + β n =\alpha^n+\beta^n = α n + β n
Quadratic equation lecture 3
Solution
α 2 − α − 1 = 0 \alpha^2-\alpha-1=0 α 2 − α − 1 = 0 ⇒ α 2 + 1 = α + 2 \Rightarrow \alpha^2+1=\alpha+2 ⇒ α 2 + 1 = α + 2
β 2 − β − 1 = 0 \beta^2-\beta-1=0 β 2 − β − 1 = 0
⇒ β 2 + 1 = β + 2 \Rightarrow \beta^2+1=\beta+2 ⇒ β 2 + 1 = β + 2
α + 2 \alpha+2 α + 2
= 1 + 5 2 + 2 =\frac{1+\sqrt{5}}{2}+2 = 2 1 + 5 + 2 = 5 + 5 2 = 5 ( 5 + 1 2 ) =\frac{5+\sqrt{5}}{2} =\sqrt{5}(\frac{\sqrt{5}+1}{2}) = 2 5 + 5 = 5 ( 2 5 + 1 ) = 5 α =\sqrt{5}\alpha = 5 α
Quadratic equation lecture 3
Solution
β + 2 \beta+2 β + 2
1 − 5 2 + 2 \frac{1-\sqrt{5}}{2}+2 2 1 − 5 + 2
5 − 5 2 \frac{\sqrt{5}-\sqrt{5}}{2} 2 5 − 5 = 5 ( 5 − 1 2 ) =\sqrt{5}(\frac{\sqrt{5}-1}{2}) = 5 ( 2 5 − 1 ) = − 5 β =-\sqrt{5} \beta = − 5 β
α − β = 5 \alpha-\beta=\sqrt{5} α − β = 5
α + β = 1 \alpha+\beta=1 α + β = 1
Quadratic equation lecture 3
Solution
∑ n ⩾ 1 b n 10 n = ∑ n ⩾ 1 α n + β n 10 n = ∑ n ⩾ 1 α n 10 n + ∑ n ⩾ 1 β n 10 n \sum_{n \geqslant 1} \frac{b_n}{10^n}=\sum_{n \geqslant 1} \frac{\alpha^n+\beta^n}{10^n}=\sum_{n \geqslant 1} \frac{\alpha^n}{10^n}+\sum_{n \geqslant 1} \frac{\beta^n}{10^n} ∑ n ⩾ 1 1 0 n b n = ∑ n ⩾ 1 1 0 n α n + β n = ∑ n ⩾ 1 1 0 n α n + ∑ n ⩾ 1 1 0 n β n
= α / 10 1 − α / 10 + β / 10 1 − β / 10 =\frac{\alpha / 10}{1-\alpha / 10}+\frac{\beta / 10}{1-\beta / 10} = 1 − α /10 α /10 + 1 − β /10 β /10
= α 10 − α + β 10 − β =\frac{\alpha}{10-\alpha}+\frac{\beta}{10-\beta} = 10 − α α + 10 − β β = α ( 10 − β ) + β ( 10 − α ) ( 10 − α ) ( 10 − β ) =\frac{\alpha(10-\beta)+\beta(10-\alpha)}{(10-\alpha)(10-\beta)} = ( 10 − α ) ( 10 − β ) α ( 10 − β ) + β ( 10 − α )
= 10 ( α + β ) − 2 α β 100 − 10 ( α + β ) + α β =\frac{10(\alpha+\beta)-2 \alpha \beta}{100-10(\alpha+\beta)+\alpha \beta} = 100 − 10 ( α + β ) + α β 10 ( α + β ) − 2 α β = 10 + 2 100 − 10 − 1 =\frac{10+2}{100-10-1} = 100 − 10 − 1 10 + 2 = 12 89 =\frac{12}{89} = 89 12
Quadratic equation lecture 3
Solution
∑ n ⩾ 1 a n 10 n = ∑ n ⩾ 1 α n − β n ( α − β ) 10 n \sum_{n \geqslant 1} \frac{a_n}{10^n}=\sum_{n \geqslant 1} \frac{\alpha^n-\beta^n}{(\alpha-\beta) 10^n} ∑ n ⩾ 1 1 0 n a n = ∑ n ⩾ 1 ( α − β ) 1 0 n α n − β n
= 1 α − β ∑ n ⩾ 1 α n 10 n − 1 α − β ∑ n ⩾ 1 β n 10 n =\frac{1}{\alpha-\beta} \sum_{n \geqslant 1} \frac{\alpha^n}{10^n}-\frac{1}{\alpha-\beta} \sum_{n \geqslant 1} \frac{\beta^n}{10^n} = α − β 1 ∑ n ⩾ 1 1 0 n α n − α − β 1 ∑ n ⩾ 1 1 0 n β n
= 1 α − β ( α 10 − α − β 10 − β ) =\frac{1}{\alpha-\beta}(\frac{\alpha}{10-\alpha}-\frac{\beta}{10-\beta}) = α − β 1 ( 10 − α α − 10 − β β ) = 1 α − β ⋅ { 10 α − α β − 10 β + α β 100 − 10 ( α + β ) + α β } =\frac{1}{\alpha-\beta} \cdot\{\frac{10 \alpha-\alpha \beta-10 \beta+\alpha \beta}{100-10(\alpha+\beta)+\alpha \beta}\} = α − β 1 ⋅ { 100 − 10 ( α + β ) + α β 10 α − α β − 10 β + α β }
= 1 α − β ⋅ 10 ( α − β ) 100 − 10 ( α + β ) + α β =\frac{1}{\alpha-\beta} \cdot \frac{10(\alpha-\beta)}{100-10(\alpha+\beta)+\alpha \beta} = α − β 1 ⋅ 100 − 10 ( α + β ) + α β 10 ( α − β ) = 10 100 − 10 − 1 =\frac{10}{100-10-1} = 100 − 10 − 1 10 = 10 89 =\frac{10}{89} = 89 10
Quadratic equation lecture 3
Solution
b n = a n − 1 + a n + 1 b_n =a_{n-1}+a_{n+1} b n = a n − 1 + a n + 1
= α n − 1 − β n − 1 α − β + α n + 1 − β n + 1 α − β =\frac{\alpha^{n-1}-\beta^{n-1}}{\alpha-\beta}+\frac{\alpha^{n+1}-\beta^{n+1}}{\alpha-\beta} = α − β α n − 1 − β n − 1 + α − β α n + 1 − β n + 1
= α n − 1 ( α 2 + 1 ) − β n − 1 ( β 2 + 1 ) α − β =\frac{\alpha^{n-1}(\alpha^2+1)-\beta^{n-1}(\beta^2+1)}{\alpha-\beta} = α − β α n − 1 ( α 2 + 1 ) − β n − 1 ( β 2 + 1 ) = α n − 1 ( α + 2 ) − β n − 1 ( β + 2 ) α − β =\frac{\alpha^{n-1}(\alpha+2)-\beta^{n-1}(\beta+2)}{\alpha-\beta} = α − β α n − 1 ( α + 2 ) − β n − 1 ( β + 2 )
= α n − β n α − β + 2 α n − 1 − β n − 1 α − β =\frac{\alpha^n-\beta^n}{\alpha-\beta}+2 \frac{\alpha^{n-1}-\beta^{n-1}}{\alpha-\beta} = α − β α n − β n + 2 α − β α n − 1 − β n − 1
Quadratic equation lecture 3
Solution
b n = a n + 2 a n − 1 b_n=a_n+2 a_{n-1} b n = a n + 2 a n − 1 = a n − 1 + a n + 1 = a_{n-1}+a_{n+1} = a n − 1 + a n + 1
a n − 1 + a n + 1 = a n + 2 a n − 1 a_{n-1}+a_{n+1}=a_n+2 a_{n-1} a n − 1 + a n + 1 = a n + 2 a n − 1
⇒ a n + 1 = a n + a n − 1 \Rightarrow a_{n+1}=a_n+a_{n-1} ⇒ a n + 1 = a n + a n − 1
a n + 2 = a n + 1 + a n a_{n+2}=a_{n+1}+a_n a n + 2 = a n + 1 + a n = a n + a n − 1 + a n =a_n+a_{n-1}+a_n = a n + a n − 1 + a n = a n − 1 + a n − 2 + a n − 1 + a n =a_{n-1}+a_{n-2}+a_{n-1}+a_n = a n − 1 + a n − 2 + a n − 1 + a n
Quadratic equation lecture 3
Solution
a n + 2 = a 2 + a 1 + a 2 + ⋯ + a n a_{n+2}=a_2+a_1+a_2+\cdots+a_n a n + 2 = a 2 + a 1 + a 2 + ⋯ + a n
a 2 = α 2 − β 2 α − β = α + β = 1 a_2=\frac{\alpha^2-\beta^2}{\alpha-\beta}=\alpha+\beta=1 a 2 = α − β α 2 − β 2 = α + β = 1
∴ a n + 2 = 1 + a 1 + a 2 + ⋯ + a n \therefore a_{n+2}=1+a_1+a_2+\cdots+a_n ∴ a n + 2 = 1 + a 1 + a 2 + ⋯ + a n
⇒ a 1 + a 2 + ⋯ + a n = a n + 2 − 1. \Rightarrow a_1+a_2+\cdots+a_n=a_{n+2}-1. ⇒ a 1 + a 2 + ⋯ + a n = a n + 2 − 1.
Quadratic equation lecture 3
Question
For a ≠ b a \neq b a = b x 2 + a x + b = 0 x^2+ax+b=0 x 2 + a x + b = 0 x 2 + b x + a = 0 x^2+bx+a=0 x 2 + b x + a = 0
(1) 1
(2) 2
(3) -1
(4) -2
Quadratic equation lecture 3
Solution
α → a \alpha \rightarrow a α → a
α 2 + a α + b = 0 \alpha^2+a \alpha+b=0 α 2 + a α + b = 0
α 2 + b α + a = 0 \alpha^2+b \alpha+a=0 α 2 + b α + a = 0
α ( a − b ) + ( b − a ) = 0 \alpha(a-b)+(b-a)=0 ~ α ( a − b ) + ( b − a ) = 0 ⇒ α ( a − b ) = ( a − b ) \Rightarrow \alpha(a-b)=(a-b) ⇒ α ( a − b ) = ( a − b ) ⇒ α = 1 ~ \Rightarrow \alpha=1 ⇒ α = 1
1 + b + a = 0
⇒ ~ \Rightarrow ⇒
Quadratic equation lecture 3
Question
For which of the following values of b, the quadratic equations x 2 x^2 x 2 x 2 x^2 x 2
(1) i 3 i\sqrt{3} i 3
(2) 2 \sqrt{2} 2
(3) − i 3 -i\sqrt{3} − i 3
(4) − 2 -\sqrt{2} − 2
Quadratic equation lecture 3
Solution
α → \alpha \rightarrow α →
α 2 + b α − 1 = 0 \alpha^2+b \alpha-1=0 α 2 + b α − 1 = 0 α 2 + α + b = 0 \alpha^2+\alpha+b=0 α 2 + α + b = 0 α ( b − 1 ) = b + 1 \alpha(b-1)=b+1 α ( b − 1 ) = b + 1
Note b ≠ \neq =
α = b + 1 b − 1 \alpha =\frac{b+1}{b-1}\qquad α = b − 1 b + 1
x 2 + x − 1 = 0 x^2+x-1=0 x 2 + x − 1 = 0
Quadratic equation lecture 3
Solution
x 2 + x + 1 = 0 x^2+x+1=0 x 2 + x + 1 = 0
No Common Solution
α 2 = 1 − b α \alpha^2 =1-b \alpha α 2 = 1 − b α = 1 − b ( b + 1 b − 1 ) =1-b(\frac{b+1}{b-1}) = 1 − b ( b − 1 b + 1 )
= b − 1 − b 2 − b b − 1 = − ( 1 + b 2 ) b − 1 =\frac{b-1-b^2-b}{b-1}=-\frac{(1+b^2)}{b-1} = b − 1 b − 1 − b 2 − b = − b − 1 ( 1 + b 2 ) = 1 + b 2 1 − b =\frac{1+b^2}{1-b} = 1 − b 1 + b 2
( b + 1 b − 1 ) 2 (\frac{b+1}{b-1})^2 ( b − 1 b + 1 ) 2
b 2 + 2 b + 1 = ( 1 − b ) ( 1 + b 2 ) b^2+2b+1 =(1-b)(1+b^2) b 2 + 2 b + 1 = ( 1 − b ) ( 1 + b 2 )
Quadratic equation lecture 3
Solution
= 1 − b + b 2 − b 3 =1-b+b^2-b^3 = 1 − b + b 2 − b 3
b 3 + 3 b = 0 b^3+3b=0 b 3 + 3 b = 0
b ( b 2 + 3 ) = 0 b(b^2+3)=0 b ( b 2 + 3 ) = 0
b = 0 o r b 2 + 3 = 0 b=0 \quad or \quad b^2+3=0 b = 0 or b 2 + 3 = 0
b = 0 b=0\qquad b = 0 b 2 = − 3 b^2=-3\quad b 2 = − 3
b = ± 3 i b= \pm \sqrt{3} i b = ± 3 i
