Question

if the sum of the roots of the quadratic equation ax2 +bx+c=0 is equal to the sum of the square of their reciprocals ,then prove that 2a2c=c2b+b2a

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Answer 1

Answer:

HOPE YOU WILL UNDERSTAND THE ANSWER AND CONCEPT.

Answer 2

Let α and β be the roots of the equation

$ax^{2} + bx^{2} + c = 0$

therefore, α + β = -b/a and αβ = c/a ....(1)

According to the given condition,

$α + β \: = \frac{1}{ α^{2} } + \frac{1}{ β ^{2} }$

$α + β = \frac{α^{2} + β ^{2} }{ (α β) ^{2} }$

$α + β = \frac{(α + β)^{2} - 2 α + β}{ (α + β)^{2} }$

$...[since, {α}^{2} + {β}^{2} = {α + β}^{2} - 2αβ]$

$\frac{ - b}{a} = \frac{( \frac{ - b}{a} ) - 2 \frac{c}{a} }{( \frac{c}{a}) ^{2} }$

$( \frac{ - b}{a}) \: ( \frac{c}{a}) ^{2} = \frac{b ^{2} }{a ^{2} } - \frac{2c}{a}$

$\frac{ - bc ^{2} }{a ^{3} } = \frac{b ^{2} - 2ca}{ a ^{2} }$

$\frac{ - bc ^{2} }{a} = {b}^{2} - 2ca$

$- bc^{2} = {b}^{2}a - 2ca^{2}$

$2ca^{2} = {b}^{2} a + {bc}^{2}$

$i.e. \: 2 {a}^{2} c = {c}^{2} b + {b}^{2} a$