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Answer 1

Answer:

$\large{\rm{ \dfrac{-(a^2 + b^2)}{2} }}$

[Option B]

Step-by-step explanation:

As per the providing information in the given question we have,

• $\sf {Eq^n = \dfrac{1}{x+a} + \dfrac{1}{x+b} = \dfrac{1}{c} }$
• Sum of zeroes of this equation is 0.

We've been asked to calculate the product of zeroes of this equation.

In order to tackle this question, we shall first simplify the equation in order to write it in the form of a quadratic equation. As per the equation we have,

$\dashrightarrow \quad \rm { \dfrac{1}{x+a} + \dfrac{1}{x+b} = \dfrac{1}{c} }$

Taking the L.C.M in LHS and and performing simplification by using the rule of addition of fractions.

$\dashrightarrow \quad \rm { \dfrac{(x+b) + (x + a)}{(x+a)(x+b)} = \dfrac{1}{c} }$

Removing the brackets & performing addition in the brackets. And, performing multiplication in the denominator using the the multiplication over addition property.

$\dashrightarrow \quad \rm { \dfrac{2x +b + a}{x^2 + bx + ax + ab} = \dfrac{1}{c} }$

Now, using the cross multiplication method to simplify further.

$\dashrightarrow \quad \rm { 1(x^2 + bx + ax + ab )= c(2x + b+ a) }$

Performing using the the multiplication over addition property in LHS and RHS.

$\dashrightarrow \quad \rm { x^2 + bx + ax + ab = 2xc + bc+ ac }$

This equation can be written as,

$\dashrightarrow \quad \rm { x^2 + bx + ax + ab -( 2xc + bc+ ac) = 0 }$

Or,

$\dashrightarrow \quad \rm { x^2 + bx + ax - 2xc + ab - bc -ac = 0 }$

Taking x as common in bx + ax - 2xc.

$\dashrightarrow \quad \boxed{ \rm{ x^2 + x(a + b - 2c) + ab - bc -ac = 0 }}$

Now it has become a quadratic equation in which,

$\dashrightarrow \quad \boxed{\rm{ \underbrace{x^2}_{ax^2}+ \underbrace{x(a + b - 2c)}_{bx} + \underbrace{ab - bc -ac}_{c} = 0 }}$

Now, according to the question, "Sum of zeroes of this equation is 0."

Let us say that the zeros of this equation be α and β. So,

$\dashrightarrow \quad \rm { \alpha + \beta = 0 }$

We know that,

When α and β are the roots or zeros of the quadratic equation ax² + bx + c = 0, then equation will be,

$\dashrightarrow \quad \boxed{\bf { x^2 - x(\alpha + \beta) + \alpha.\beta = 0} }$

Henceforth,

$\dashrightarrow \quad \boxed{\bf{ \underbrace{x^2}_{x^2} - \Bigg ( + \underbrace{x(a + b - 2c)}_{ x(\alpha + \beta) } \Bigg ) + \underbrace{ab - bc -ac}_{ \alpha.\beta} = 0 }} \dots ( 1 )$

From ( 1 ) and the given information in the question, we can conclude that,

$\dashrightarrow \quad \rm { \alpha + \beta = 0 }$

Substitute the value of α and β as per the equation ( 1 ).

$\dashrightarrow \quad \rm { a + b - 2c = 0 }$

From this equation, we'll find the value of c suggested we will write the value of c in the terms of a and b.

With the help of transposition method, solving it further.

$\dashrightarrow \quad \rm { - 2c = 0 -a - b}$

Transposing -2 from LHS to RHS. Its arithmetic operator will get changed.

$\dashrightarrow \quad \rm { c = \dfrac{-a - b}{-2}}$

Now, this can be written as,

$\dashrightarrow \quad \rm { c = \dfrac{\cancel{-}(a + b)}{\cancel{-} 2}}$

$\dashrightarrow \quad \underline{\boxed{ \bf { c = \dfrac{a + b}{2}}}}$

Now, also from ( 1 ), we can conclude that,

$\dashrightarrow \quad \rm { Product_{(Of\; Zeroes)} = ab - bc - ac }$

Taking b as common in (ab - bc).

$\dashrightarrow \quad \rm { \alpha.\beta = b(a - c) - ac }$

Now, substitute the value of c and simply it further.

$\dashrightarrow \quad \rm { \alpha.\beta = b\Bigg ( a - \dfrac{a +b}{2} \Bigg ) - a\Bigg (\dfrac{a +b}{2} \Bigg ) }$

Performing addition and multiplication.

$\dashrightarrow \quad \rm { \alpha.\beta = b\Bigg ( \dfrac{2a -a-b}{2} \Bigg ) - \dfrac{a^2 + ab}{2} }$

$\dashrightarrow \quad \rm { \alpha.\beta = b\Bigg ( \dfrac{a - b}{2} \Bigg ) - \dfrac{a^2 + ab}{2} }$

$\dashrightarrow \quad \rm { \alpha.\beta = \dfrac{ab - b^2}{2} - \dfrac{a^2 + ab}{2} }$

$\dashrightarrow \quad \rm { \alpha.\beta = \dfrac{ab - b^2}{2} - \dfrac{a^2 + ab}{2} }$

$\dashrightarrow \quad \rm { \alpha.\beta = \dfrac{ab - b^2 - (a^2 + ab)}{2} }$

$\dashrightarrow \quad \rm { \alpha.\beta = \dfrac{\cancel{ab} - b^2 - a^2\cancel{ - ab}}{2} }$

$\dashrightarrow \quad \rm { \alpha.\beta = \dfrac{- b^2 - a^2}{2} }$

This can also be written as,

$\dashrightarrow \quad \underline{\boxed{\bf { \alpha.\beta = \dfrac{-(a^2 + b^2 )}{2}}} }$

∴ The product of zeroes of the given equation is $\bf \dfrac{-(a^2 + b^2 )}{2}$.

Option B is the correct option.

$\rule{200}2$