Question

if alpha and beta are the zeros of the polynomial 2xsquare -4x+5 find the value of(a) alpha square + beta square (b) alpha - beta the whole square (c) 1 by alpha+ 1 by beta (d) 1 by alpha square + 1 by beta square (e) alpha cube + beta cube​

Answer 1

$\Large{\underline{\underline{\mathfrak{\bf{Question}}}}}$

If α and β are the zeros of the polynomial 2x² - 4x + 5 = 0 .find the value of :-

(a) α² - β²

(b) (α - β)²

(c) 1/α + 1/β

(d) 1/α² + 1/β²

$\Large{\underline{\underline{\mathfrak{\bf{Solution}}}}}$

$\Large{\underline{\mathfrak{\bf{Given}}}}$

• polynomial 2x² - 4x + 5 = 0
• α and β are the zeros.

$\Large{\underline{\mathfrak{\bf{Find}}}}$

• (α² - β² )
• (α - β)²
• ( 1/α + 1/β)
• (1/α² + 1/β²)

$\Large{\underline{\underline{\mathfrak{\bf{Explanation}}}}}$

We know,

$\small\boxed{\sf{\pink{\:Sum\:of\:zeroes\:=\:\dfrac{-(coefficient\:of\:x)}{(coefficient\:of\:x^2)}}}} \\ \\ \mapsto\sf{\:(\alpha\:+\:\beta)\:=\:\dfrac{-(-4)}{2}} \\ \\ \mapsto\sf{\:(\alpha\:+\:\beta)\:=\:\dfrac{\cancel{4}}{\cancel{2}}} \\ \\ \mapsto\sf{\:(\alpha\:+\:\beta)\:=\:2.......(1)}$

Again,

$\small\boxed{\sf{\pink{\:product\:of\:zeroes\:=\:\dfrac{(constant\:part)}{(coefficient\:of\:x^2)}}}} \\ \\ \mapsto\sf{\:(\alpha . \beta)\:=\:\dfrac{5}{2}.....(2)}$

We know,

$\small\boxed{\sf{\orange{\:(\alpha\:-\:\beta)\:=\:\sqrt{(\alpha\:+\:\beta)^2-4\alpha.\beta}}}} \\ \\ \small\sf{\:\:\:\:\:\:keep\:value\:by\:equ(1)\:and\:(2)} \\ \\ \mapsto\sf{\:(\alpha\:-\:\beta)\:=\:\sqrt{(2)^2-4\times \dfrac{5}{2}}} \\ \\ \mapsto\sf{\:(\alpha\:-\:\beta)\:=\:\sqrt{(4-2\times 5}} \\ \\ \mapsto\sf{\:(\alpha\:-\:\beta)\:=\:\sqrt{4-10}} \\ \\ \mapsto\sf{\:(\alpha\:-\:\beta)\:=\:\sqrt{-6}} \\ \\ \mapsto\sf{\:(\alpha\:-\:\beta)\:=\:i\sqrt{6}.....(3)}$

Add equation (1) and (3) ,

$\mapsto\sf{\:2\alpha\:=\:(2+i\sqrt{6})} \\ \\ \boxed{\sf{\:\alpha\:=\:\dfrac{(2+i\sqrt{6})}{2}}}$

Again, keep value of α in equ (1)

$\mapsto\sf{\:\dfrac{(2+i\sqrt{6})}{2}\:+\:\beta\:=\:2} \\ \\ \mapsto\sf{\:\beta\:=\:2\:-\:\dfrac{(2+i\sqrt{6})}{2}} \\ \\ \mapsto\sf{\:\beta\:=\:\dfrac{4-(2+i\sqrt{6})}{2}} \\ \\ \boxed{\sf{\:\beta\:=\:\dfrac{(2-i\sqrt{6})}{2}}}$

First:-

• (α² - β² )

keep value of α and β

$\mapsto\sf{\:\left(\dfrac{(2+i\sqrt{6})}{2}\right)^2\:-\:\left(dfrac{(2-i\sqrt{6})}{2}\right)^2} \\ \\ \mapsto\sf{\:\left(\dfrac{(2+i\sqrt{6})^2-(2-i\sqrt{6})^2}{4}\right)} \\ \\ \mapsto\sf{\:\dfrac{2^2+i^2\times (\sqrt{6})^2+4i\sqrt{6}-2^2-i^2\times (\sqrt{6})^2+4i\sqrt{6}}{4}} \\ \\ \mapsto\sf{\:\dfrac{(\cancel{8}i\sqrt{6}}{\cancel{4}}} \\ \\ \mapsto\sf{\red{\:2i\sqrt{6}\:\:\:\:\:\:\:\:\:Ans}}$

Or,

• (α² + β² ).

keep value of α and β

$\mapsto\sf{\:\left(\dfrac{(2+i\sqrt{6})}{2}\right)^2\:+\:\left(\dfrac{(2-i\sqrt{6})}{2}\right)^2} \\ \\ \mapsto\sf{\:\left(\dfrac{(2+i\sqrt{6})^2+(2-i\sqrt{6})^2}{4}\right)} \\ \\ \mapsto\sf{\:\dfrac{2^2+i^2\times (\sqrt{6})^2+4i\sqrt{6}+2^2+i^2\times (\sqrt{6})^2-4i\sqrt{6}}{4}} \\ \\ \mapsto\sf{\:\dfrac{(4+4+(-1)\times 6+ (-1)\times 6}{4}} \\ \\ \mapsto\sf{\:\dfrac{(8-12)}{4}} \\ \\ \mapsto\sf{\:\dfrac{\cancel{-4}}{\cancel{4}}} \\ \\ \mapsto\sf{\red{\:-1\:\:\:\:\:\:\:\:\:Ans}}$

Second:-

• (α - β)²

keep value of α and β

$\mapsto\sf{\:\left(\dfrac{(2+i\sqrt{6})}{2}\:-\:\dfrac{(2-i\sqrt{6})}{2}\right)^2} \\ \\ \mapsto\sf{\:\left(\dfrac{2+i\sqrt{6}-2+i\sqrt{6}}{2}\right)^2} \\ \\ \mapsto\sf{\:\dfrac{(2i\sqrt{6})^2}{4}} \\ \\ \mapsto\sf{\:\dfrac{2^2\times i^2\times (\sqrt{6})^2}{4}} \\ \\ \mapsto\sf{\:\dfrac{\cancel{4}\times -1\times 6}{\cancel{4}}} \\ \\ \mapsto\sf{\pink{\:-6\:\:\:\:\:\:\:\:\:Ans}}$

Third:-

• ( 1/α + 1/β)

keep value of α and β

$\mapsto\sf{\:\dfrac{\alpha+\beta}{\alpha.\beta}} \\ \\ \small\sf{\:\:\:\:\:keep\:value\:by\:equ(1)\:and\:(2)} \\ \\ \mapsto\sf{\:\dfrac{2}{\dfrac{5}{2}}} \\ \\ \mapsto\sf{\:\dfrac{2\times2}{5}} \\ \\ \mapsto\sf{\orange{\:\dfrac{4}{5}\:\:\:\:\:\:\:\:\:\:\:Ans}}$

Forth:-

• (1/α² + 1/β²)

$\mapsto\sf{\:\dfrac{(\alpha^2+\beta^2)}{\alpha^2.\beta^2}} \\ \\ \mapsto\sf{\:\dfrac{-1}{(\dfrac{5}{2})^2}} \\ \\ \mapsto\sf{\:\dfrac{-1\times 4}{25}} \\ \\ \mapsto\sf{\pink{\:\dfrac{-4}{25}\:\:\:\:\:\:\:\:\:\:Ans}}$

$\Large{\underline{\mathfrak{\bf{\orange{Using\:Identity}}}}}$

• (x+y)² = x² + y² +2xy
• (x-y)² = x² + y² - 2xy
• (i)² = -1

Answer 2

Step-by-step explanation:

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