Question

if α, β be the roots of ax^2-x+a=0 find the equation whose roots are α/β, β/α.​

Answer 1

Answer:

The required quadratic equation is $\displaystyle{\sf\:a^2\:x^2\:+\:2a^2x\:=\:0}$.

Step-by-step-explanation:

The given quadratic equation is $\displaystyle{\sf\:ax^2\:-\:x\:+\:a\:=\:0}$

The roots of the given quadratic equation are $\displaystyle{\sf\:\alpha\:\&\:\beta}$.

We have to find a quadratic equation whose roots are $\displaystyle{\sf\:\dfrac{\alpha}{\beta}\:\&\:\dfrac{\beta}{\alpha}}$

Now,

$\displaystyle{\sf\:ax^2\:-\:x\:+\:a\:=\:0}$

Comparing with $\displaystyle{\sf\:ax^2\:+\:bx\:+\:c\:=\:0}$, we get,

$\displaystyle{\bullet\:\sf\:a\:=\:a}$

$\displaystyle{\bullet\:\sf\:b\:=\:-\:1}$

$\displaystyle{\bullet\:\sf\:c\:=\:a}$

Now, we know that,

$\displaystyle{\pink{\sf\:Sum\:of\:roots\:(\:\alpha\:+\:\beta\:)\:=\:-\:\dfrac{b}{a}}}$

$\displaystyle{\implies\sf\:\alpha\:+\:\beta\:=\:-\:\left(\:\dfrac{-\:1}{a}\:\right)}$

$\displaystyle{\implies\sf\:\alpha\:+\:\beta\:=\:\dfrac{1}{a}\:\:\:-\:-\:-\:(\:1\:)}$

Now,

$\displaystyle{\pink{\sf\:Product\:of\:roots\:(\:\alpha\:.\:\beta\:)\:=\:\dfrac{c}{a}}}$

$\displaystyle{\implies\sf\:\alpha\:.\:\beta\:=\:\cancel{\dfrac{a}{a}}}$

$\displaystyle{\implies\sf\:\alpha\:.\:\beta\:=\:1\:\:\:-\:-\:-\:(\:2\:)}$

We have to find a quadratic equation whose roots are $\displaystyle{\sf\:\dfrac{\alpha}{\beta}\:\&\:\dfrac{\beta}{\alpha}}$

The quadratic equation is in form

$\displaystyle{\pink{\sf\:x^2\:-\:(\:\alpha\:+\:\beta\:)\:x\:+\:\alpha\:.\:\beta\:=\:0}}$

$\displaystyle{\implies\sf\:x^2\:-\:\left(\:\dfrac{\alpha}{\beta}\:+\:\dfrac{\beta}{\alpha}\:\right)\:x\:+\:\dfrac{\cancel{\alpha}}{\cancel{\beta}}\:\times\:\dfrac{\cancel{\beta}}{\cancel{\alpha}}\:=\:0}$

$\displaystyle{\implies\sf\:x^2\:-\:\left(\:\dfrac{\alpha^2\:+\:\beta^2}{\alpha\:.\:\beta}\:\right)\:x\:+\:1\:=\:0\:\:\:-\:-\:-\:[\:From\:(\:2\:)\:]}$

$\displaystyle{\implies\sf\:\left[\:\dfrac{(\:\alpha\:+\:\beta\:)^2\:-\:2\:\alpha\:\beta}{\alpha\:.\:\beta}\:\right]\:x\:+\:1\:=\:0\:\:\:-\:-\:-\:[\:(\:a\:+\:b\:)^2\:=\:a^2\:+\:2ab\:+\:b^2\:]}$

$\displaystyle{\implies\sf\:x^2\:-\:\left[\:\dfrac{\left(\:\dfrac{1}{a}\:\right)^2\:-\:2\:\times\:1}{1}\:\right]\:x\:+\:1\:=\:0\:\:\:-\:-\:-\:[\:From\:(\:1\:)\:]}$

$\displaystyle{\implies\sf\:x^2\:-\:\left(\:\dfrac{1}{a^2}\:-\:2\:\right)\:x\:+\:1\:=\:0}$

$\displaystyle{\implies\sf\:x^2\:-\:\left(\:\dfrac{1\:-\:2a^2}{a^2}\:\right)\:x\:+\:1\:=\:0}$

$\displaystyle{\implies\sf\:a^2\:x^2\:-\:\cancel{1}\:+\:2a^2x\:+\:\cancel{1}\:=\:0}$

$\displaystyle{\implies\underline{\boxed{\red{\sf\:a^2\:x^2\:+\:2a^2x\:=\:0}}}}$

The required quadratic equation is $\displaystyle{\sf\:a^2\:x^2\:+\:2a^2x\:=\:0}$.

Answer 2

$\huge\pink{\boxed{\blue{\boxed{ \purple{ \boxed{{\pink{Answer}}}}}}}} \\ \large\pink{\boxed{\blue{\boxed{ \purple{ \boxed{{\pink{Your~answer↓}}}}}}}}$

$Since \: \alpha \: and \: \beta \: are \: the \: roots \: of \: the \: equation \: {ax}^{2} - x + a = 0.$

On comparing it with the equation ax² + bx + c = 0, we get

♡ a = a

♡ b = - 1

♡ c = a

$Sum \: of \: zeroes \: = \alpha + \beta = - \frac{b}{a} = - \frac{ -1 }{a} = \frac{1}{a}$

$\small\bold\red{Product \: of \: zeroes \: = \alpha \beta \: = \frac{c}{a} = \frac{a}{a} = 1}$

Now, we have to find a quadratic equation whose roots are

$\frac{ \alpha }{ \beta } \: and \: \frac{ \beta }{ \alpha }$

$\small\bold\red{Sum \: of \: zeroes \: = } \\ \small\bold\red{ \frac{ \alpha }{ \beta } + \frac{ \beta }{ \alpha } } \\ \small\bold\red{ = \frac{ { \alpha }^{2} + { \beta }^{2} }{ \alpha \beta } } \\ \small\bold\red{ = \frac{ {( \alpha + \beta )}^{2} - 2 \alpha \beta }{ \alpha \beta } } \\ \small\bold\red{ = \frac{ {( \frac{1}{a} )}^{2} - 2}{1} } \\ \small\bold\red{ = \frac{1 - {2a}^{2} }{ {a}^{2} } }$

$\small\bold\red{Product \: of \: zeroes \: = } \\ \small\bold\red{ = \frac{ \alpha }{ \beta } \times \frac{ \beta }{ \alpha } } \\ \small\bold\red{ = 1}$

Now the required quadratic equation is

$\small\bold\red{ {x}^{2} - ( \frac{ \alpha }{ \beta } + \frac{ \beta }{ \alpha } )x + ( \frac{ \alpha }{ \beta } \times \frac{ \beta }{ \alpha }) = 0 } \\ \small\bold\red{ = > \: {x}^{2} - ( \frac{1 - {2a}^{2}}{ {a}^{2} } ) x + 1 = 0} \\ \small\bold\red{}\small\bold\red{ = > \: {a}^{2} {x}^{2} - x + {2a}^{2} x + {a}^{2} = 0}$

$\huge \fcolorbox{black}{cyan}{♛Hope it helps U♛}$

$\red{\textsf{(✪MARK BRAINLIEST✿)}} \\ \huge\fcolorbox{aqua}{aqua}{\red{FolloW ME}}$