Question

If the sum of the roots of the equation ax + bx + c = 0 is equal to the sum
of the squares of their reciprocals
, then prove that bc?, ca?, ab? are in A.P.​

Answer 1

Answer

If the sum of the roots of the quadratic equation ax2+bx+c=0 is equal to the sum of the squares of their reciprocals then b2/ac + bc/a2 is equal to.

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Answer 2

Correct Question is

• If the sum of the roots of the equation ax² + bx + c = 0 is equal to sum of their reciprocals, prove that

$\sf \: {bc}^{2}, \: {ca}^{2}, \: {ab}^{2} \: are \: in \: AP$

Answer

Concept used :-

We know that,

• If 'α' and 'β' are two zeroes of equation ax² + bx + c = 0, then

$\boxed{\purple{\tt Sum\ of\ the\ zeroes=\frac{-b}{a}}}$

OR

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

And

$\boxed{\purple{\tt Product\ of\ the\ zeroes=\frac{c}{a}}}$

OR

$\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$

$\large\underline\blue{\bold{ \sf \: Given - }}$

• The sum of the roots of the equation ax² + bx + c = 0 is equal to sum of their reciprocals.

$\large\underline\blue{\bold{ \sf \: Tᴏ \: Fɪɴᴅ - }}$

$\bull \: \sf \: {bc}^{2}, \: {ca}^{2}, \: {ab}^{2} \: are \: in \: AP$

$\large \boxed{ \red{ \underline{ \bf \: Cᴀʟᴄᴜʟᴀᴛɪᴏɴ :}}}$

Let 'α' and 'β' are two zeroes of equation ax² + bx + c = 0,

So,

$\sf \: \alpha + \beta = - \dfrac{b}{a}$

$\sf \: \alpha \beta = \dfrac{c}{a}$

Now,

• According to statement,

we have

$\rm :\implies\: \alpha + \beta = \dfrac{1}{ { \alpha }^{2} } + \dfrac{1}{ { \beta }^{2} }$

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

$\rm :\implies\: - \dfrac{b}{a} = (( { \alpha + \beta )}^{2} - 2 \alpha \beta ) \times \dfrac{ {a}^{2} }{ {c}^{2} }$

$\rm :\implies\: - {bc}^{2} = ( \dfrac{ {b}^{2} }{ {a}^{2} } - 2\dfrac{c}{a} ) \times {a}^{3}$

$\rm :\implies\: - {bc}^{2} = (\dfrac{ {b}^{2} - 2ac}{ {a}^{2} } ) \times {a}^{3}$

$\rm :\implies\: - {bc}^{2} = ( {b}^{2} - 2ac) \times a$

$\rm :\implies\: - {bc}^{2} = a{b}^{2} - 2c {a}^{2}$

$\rm :\implies\: {2ac}^{2} = {bc}^{2} + {ab}^{2}$

$\rm :\implies\: \boxed{ \pink{ \bf \: \: {bc}^{2}, \: {ca}^{2}, \: {ab}^{2} \: are \: in \: AP \: }}$

$\large{\boxed{\boxed{\bf{Hence, Proved}}}}$