Question

if sum of roots equal to the sum of squares of their reciprocals of a quadratic equation
ax2 + bx + c = 0 then prove that bc² ca², ab² are in AP​

Answer 1

Answer:

bc², ca², ab² are in AP.

Step-by-step-explanation:

The given quadratic equation is ax² + bx + c = 0.

Let the roots of the quadratic equation be $\displaystyle{\sf\:\alpha\:\&\:\beta}$

Now, we know that,

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

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

We have given that,

The sum of roots of the quadratic equation is equal to sum of squares of their reciprocals.

$\displaystyle{\therefore\:\sf\:\alpha\:+\:\beta\:=\:\left(\:\dfrac{1}{\alpha}\:\right)^2\:+\:\left(\:\dfrac{1}{\beta}\:\right)^2}$

$\displaystyle{\implies\sf\:-\:\dfrac{b}{a}\:=\:\dfrac{1}{\alpha^2}\:+\:\dfrac{1}{\beta^2}}$

$\displaystyle{\implies\sf\:-\:\dfrac{b}{a}\:=\:\dfrac{\alpha^2\:+\:\beta^2}{\alpha^2\:.\:\beta^2}}$

$\displaystyle{\implies\sf\:-\:\dfrac{b}{a}\:=\:\dfrac{(\:\alpha\:+\:\beta\:)^2\:-\:2\:\alpha\:\beta}{\alpha\:.\:\beta\:.\:\alpha\:.\:\beta}}$

$\displaystyle{\implies\sf\:-\:\dfrac{b}{a}\:=\:\dfrac{\left(\:-\:\dfrac{b}{a}\:\right)^2\:-\:2\:\dfrac{c}{a}}{\dfrac{c}{a}\:\times\:\dfrac{c}{a}}}$

$\displaystyle{\implies\sf\:-\:\dfrac{b}{a}\:=\:\dfrac{\dfrac{b^2}{a^2}\:-\:\dfrac{2c}{a}}{\dfrac{c^2}{a^2}}}$

$\displaystyle{\implies\sf\:-\:\dfrac{b}{a}\:=\:\dfrac{\dfrac{ab^2\:-\:2ca^2}{a^3}}{\dfrac{c^2}{a^2}}}$

$\displaystyle{\implies\sf\:-\:\dfrac{b}{a}\:=\:\dfrac{ab^2\:-\:2ca^2}{a^3}\:\times\:\dfrac{a^2}{c^2}}$

$\displaystyle{\implies\sf\:-\:\dfrac{b}{a}\:=\:\dfrac{ab^2\:-\:2ca^2}{\cancel{a^2}\:\times\:a}\:\times\:\dfrac{\cancel{a^2}}{c^2}}$

$\displaystyle{\implies\sf\:-\:\dfrac{b}{a}\:=\:\dfrac{ab^2\:-\:2ca^2}{a}\:\times\:\dfrac{1}{c^2}}$

$\displaystyle{\implies\sf\:\dfrac{-\:b}{a}\:=\:\dfrac{ab^2\:-\:2ca^2}{ac^2}}$

$\displaystyle{\implies\sf\:\dfrac{-\:b}{\cancel{a}}\:\times\:\cancel{a}\:c^2\:=\:ab^2\:-\:2ca^2}$

$\displaystyle{\implies\sf\:-\:bc^2\:=\:ab^2\:-\:2ca^2}$

$\displaystyle{\implies\sf\:-\:bc^2\:=\:ab^2\:-\:ca^2\:-\:ca^2}$

$\displaystyle{\implies\boxed{\red{\sf\:ca^2\:-\:bc^2\:=\:ab^2\:-\:ca^2\:}}}$

∴ bc², ca², ab² are in AP.

Hence proved!