Solve these 2 questions with steps......

1.	Solve these 2 questions with steps......
Answer» $\Large{\underline{\underline{\mathfrak{\bf{Solution}}}}}$ For first question:- $\Large{\underline{\mathfrak{\bf{Given}}}}$ Equation, f(X) = x² - PX +q α and β are ZEROES . $\Large{\underline{\mathfrak{\bf{Find}}}}$ α² + β² 1/α + 1/β $\Large{\underline{\underline{\mathfrak{\bf{Explanation}}}}}$ we know, $\small\boxed{\sf{\:Sum\:of\:zeroes\:=\:\dfrac{-(coefficient\:of\:x)}{(coefficient\:of\:x^2)}}}$ $:\mapsto\sf{\:(\alpha + \beta)\:=\:\dfrac{-(-p)}{1}\:=\:p.........(1)}$ Again, $\small\boxed{\sf{\:product\:of\:zeroes\:=\:\dfrac{(constant\:part)}{(coefficient\:of\:x^2)}}}$ $:\mapsto\sf{\:(\alpha.\beta)\:=\:\dfrac{q}{1}\:=\:q........(2)}$ Now, first calculate , $:\mapsto\sf{\:(\alpha^2 + \beta^2)} \\ \\ :\mapsto\sf{\:(\alpha+\beta)^2-2\alpha.\beta} \\ \\ \small\sf{\:\:\:\:keep\:value\:by\:equ(1)\:and\:equ(2)} \\ \\ :\mapsto\sf{\:(p^2-2.q)\:\:\:\:\:\:\:Ans}$ Now, second calculate, $:\mapsto\sf{\:\dfrac{1}{\alpha}+\dfrac{1}{\beta}} \\ \\ :\mapsto\sf{\:\dfrac{(\alpha+\beta)}{\alpha.\beta}} \\ \\ \small\sf{\:\:\:\:\:keep\:value\:by\:equ(1)\:and\:equ(2)} \\ \\ :\mapsto\sf{\:\dfrac{p}{q}\:\:\:\:\:\:Ans}$ ______________________ For , Second question, $\Large{\underline{\mathfrak{\bf{Given}}}}$ equation, f(x)=ax² + bx + c α and β are zeroes . $\Large{\underline{\mathfrak{\bf{Find}}}}$ α⁴ + β⁴ α²/β² + β²/α² $\Large{\underline{\underline{\mathfrak{\bf{Explanation}}}}}$ we know, $\small\boxed{\sf{\:Sum\:of\:zeroes\:=\:\dfrac{-(coefficient\:of\:x)}{(coefficient\:of\:x^2)}}}$ $:\mapsto\sf{\:(\alpha + \beta)\:=\:\dfrac{-(b)}{a}.....(3)}$ Again, $\small\boxed{\sf{\:product\:of\:zeroes\:=\:\dfrac{(constant\:part)}{(coefficient\:of\:x^2)}}}$ $:\mapsto\sf{\:(\alpha.\beta)\:=\:\dfrac{c}{a}\:=\:q........(4)}$ Now, $:\mapsto\sf{\:(\alpha^2 + \beta^2)} \\ \\ :\mapsto\sf{\:(\alpha+\beta)^2-2\alpha.\beta} \\ \\ \small\sf{\:\:\:\:keep\:value\:by\:equ(3)\:and\:equ(4)} \\ \\ :\mapsto\sf{\:(\dfrac{-b}{a})^2-2.\dfrac{c}{a}} \\ \\ :\mapsto\sf{\:\dfrac{(b^2-2ac)}{a^2}.......(5)}$ Now, first calculate, $:\mapsto\sf{\:(\alpha^4+\beta^4)} \\ \\ :\mapsto\sf{\:(\alpha^2+\beta^2)^2-2.\alpha^2.\beta^2} \\ \\ \small\sf{\:keep\:value\:by\:equ(4)\:and\:equ(5)} \\ \\ :\mapsto\sf{\:(\dfrac{b^2-2ac}{a^2})^2-2.(\dfrac{c}{a})^2} \\ \\ :\mapsto\sf{\:\dfrac{(b^2-2ac)^2-2c^2}{a^2}\:\:\:Ans}$ Second, calculate $:\mapsto\sf{\:\dfrac{\alpha^2}{\beta^2}+\dfrac{\beta^2}{\alpha^2}} \\ \\ :\mapsto\sf{\:\dfrac{(\alpha^4+\beta^4)}{\alpha^2.\beta^2}} \\ \\ \small\sf{\:\:\:\:(\alpha^4+\beta^4)\:=\:\dfrac{(b^2-2ac)^2-2c^2}{a^2}} \\ \\ :\mapsto\sf{\:\dfrac{\dfrac{(b^2-2ac)^2-2c^2}{a^2}}{(\dfrac{c}{a})^2}} \\ \\ :\mapsto\sf{\:\dfrac{(b^2-2ac)^2-2c^2}{a^2}\times \dfrac{a^2}{c^2}} \\ \\ :\mapsto\sf{\:\dfrac{(b^2-2ac)^2}{c^2}\:\:\:\:\:Ans}$

Answer»

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

For first question:-

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

Equation, f(X) = x² - PX +q
α and β are ZEROES .

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

α² + β²
1/α + 1/β

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

we know,

$\small\boxed{\sf{\:Sum\:of\:zeroes\:=\:\dfrac{-(coefficient\:of\:x)}{(coefficient\:of\:x^2)}}}$

$:\mapsto\sf{\:(\alpha + \beta)\:=\:\dfrac{-(-p)}{1}\:=\:p.........(1)}$

Again,

$\small\boxed{\sf{\:product\:of\:zeroes\:=\:\dfrac{(constant\:part)}{(coefficient\:of\:x^2)}}}$

$:\mapsto\sf{\:(\alpha.\beta)\:=\:\dfrac{q}{1}\:=\:q........(2)}$

Now, first calculate ,

$:\mapsto\sf{\:(\alpha^2 + \beta^2)} \\ \\ :\mapsto\sf{\:(\alpha+\beta)^2-2\alpha.\beta} \\ \\ \small\sf{\:\:\:\:keep\:value\:by\:equ(1)\:and\:equ(2)} \\ \\ :\mapsto\sf{\:(p^2-2.q)\:\:\:\:\:\:\:Ans}$

Now, second calculate,

$:\mapsto\sf{\:\dfrac{1}{\alpha}+\dfrac{1}{\beta}} \\ \\ :\mapsto\sf{\:\dfrac{(\alpha+\beta)}{\alpha.\beta}} \\ \\ \small\sf{\:\:\:\:\:keep\:value\:by\:equ(1)\:and\:equ(2)} \\ \\ :\mapsto\sf{\:\dfrac{p}{q}\:\:\:\:\:\:Ans}$

______________________

For , Second question,

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

equation, f(x)=ax² + bx + c
α and β are zeroes .

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

α⁴ + β⁴
α²/β² + β²/α²

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

we know,

$\small\boxed{\sf{\:Sum\:of\:zeroes\:=\:\dfrac{-(coefficient\:of\:x)}{(coefficient\:of\:x^2)}}}$

$:\mapsto\sf{\:(\alpha + \beta)\:=\:\dfrac{-(b)}{a}.....(3)}$

Again,

$\small\boxed{\sf{\:product\:of\:zeroes\:=\:\dfrac{(constant\:part)}{(coefficient\:of\:x^2)}}}$

$:\mapsto\sf{\:(\alpha.\beta)\:=\:\dfrac{c}{a}\:=\:q........(4)}$

Now,

$:\mapsto\sf{\:(\alpha^2 + \beta^2)} \\ \\ :\mapsto\sf{\:(\alpha+\beta)^2-2\alpha.\beta} \\ \\ \small\sf{\:\:\:\:keep\:value\:by\:equ(3)\:and\:equ(4)} \\ \\ :\mapsto\sf{\:(\dfrac{-b}{a})^2-2.\dfrac{c}{a}} \\ \\ :\mapsto\sf{\:\dfrac{(b^2-2ac)}{a^2}.......(5)}$

Now, first calculate,

$:\mapsto\sf{\:(\alpha^4+\beta^4)} \\ \\ :\mapsto\sf{\:(\alpha^2+\beta^2)^2-2.\alpha^2.\beta^2} \\ \\ \small\sf{\:keep\:value\:by\:equ(4)\:and\:equ(5)} \\ \\ :\mapsto\sf{\:(\dfrac{b^2-2ac}{a^2})^2-2.(\dfrac{c}{a})^2} \\ \\ :\mapsto\sf{\:\dfrac{(b^2-2ac)^2-2c^2}{a^2}\:\:\:Ans}$

Second, calculate

$:\mapsto\sf{\:\dfrac{\alpha^2}{\beta^2}+\dfrac{\beta^2}{\alpha^2}} \\ \\ :\mapsto\sf{\:\dfrac{(\alpha^4+\beta^4)}{\alpha^2.\beta^2}} \\ \\ \small\sf{\:\:\:\:(\alpha^4+\beta^4)\:=\:\dfrac{(b^2-2ac)^2-2c^2}{a^2}} \\ \\ :\mapsto\sf{\:\dfrac{\dfrac{(b^2-2ac)^2-2c^2}{a^2}}{(\dfrac{c}{a})^2}} \\ \\ :\mapsto\sf{\:\dfrac{(b^2-2ac)^2-2c^2}{a^2}\times \dfrac{a^2}{c^2}} \\ \\ :\mapsto\sf{\:\dfrac{(b^2-2ac)^2}{c^2}\:\:\:\:\:Ans}$