Question

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If the sum of the roots of the quadratic equation ax + bx + c = 0 is equal to the
sum of the squares of their reciprocals, then prove that 2ac = c+b + b-a.​

Answer 1

2a²c = ab² + bc²

Step-by-step explanation:

The given quadratic equation is

ax² + bx + c = 0 ..... (1)

Let α, β be the roots of (1) no. equation.

Then by the relation between roots and coefficients, we get

α + β = - b/a ..... (2)

αβ = c/a ..... (3)

& the sum of the squares of their reciprocals is

(1/α² + 1/β²).

By the given condition,

α + β = 1/α² + 1/β²

or, - b/a = (1/α + 1/β)² - 2 (1/α * 1/β)

or, - b/a = {(β + α)/(αβ)}² - 2/(αβ)

or, - b/a = {(- b/a) / (c/a)}² - 2/(c/a)

or, - b/a = (- b/c)² - 2 (a/c)

or, - b/a = b²/c² - 2a/c

or, - b/a = (b² - 2ac)/c²

or, - bc² = ab² - 2a²c

or, 2a²c = ab² + bc²

This is the required relation.

Related question:

Find zeroes of this polynomial and verify the relationship between the zeroes and their coefficients. - https://brainly.in/question/9601995

Answer 2

The condition  2 a² c = c² b + b²a is proved

Step-by-step explanation:

Given as :

For quadratic equation in the form of

a x² + b x + c = 0

The sum of the roots of the quadratic equation = sum of the squares of their reciprocals

Let The two roots of equation = m , n

Let The reciprocal of two roots = $\dfrac{1}{m}$  ,  $\dfrac{1}{n}$

To prove : 2 a² c = c² b + b² a

According to question

∵ For quadratic equation a x² + b x + c = 0

The sum of roots of equation = m + n = $\dfrac{-b}{a}$                ........1

The products of roots of equation = m n =  $\dfrac{c}{a}$             .........2

Again

The reciprocal of roots are $\dfrac{1}{m^{2} }$  ,  $\dfrac{1}{n^{2} }$

So, The sum of reciprocal of the roots =  $\dfrac{1}{m^{2} }$  +  $\dfrac{1}{n^{2} }$

Now, A/Q ,

sum of the roots  = sum of the squares of their reciprocals

i.e  m + n = $\dfrac{1}{m^{2} }$  +  $\dfrac{1}{n^{2} }$

Or, m + n = $\dfrac{m^{2}+n^{2} }{m^{2} n^{2} }$

Or, ( m + n ) × ( m n )²  = m² + n²

Or, ( m + n ) × ( m n )²  = ( m + n )² - 2 m n

From eq 1 and eq 2, substitute the values , we get

( $\dfrac{-b}{a}$  ) × (  $\dfrac{c}{a}$  )²  = ( $\dfrac{-b}{a}$  )² - 2 (  $\dfrac{c}{a}$  )

Or, $\dfrac{- b c^{2} }{a^{^{3}}}$ =   $\dfrac{b^{2} }{a^{2} }$  - $\dfrac{2c}{a}$

Taking LCM

$\dfrac{- b c^{2} }{a^{^{3}}}$  =   $\dfrac{b^{2}-2ca}{a^{2}}}$

Or,  - b c²  =  ( b² - 2 c a )  × $\dfrac{a^{3} }{a^{2} }$

Or,    - b c²  =  ( b² - 2 c a )  × a

or,  - b c²  =   b² a - 2 c a²

Or, 2 c a² =   b² a + - b c²

So, Left hand side = right hand side, so the condition for the given statement is proved

Hence, 2 a² c = c² b + b² a is proved . Answer