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Q.

If α and β are the zeros of the quadratic polynomial f(x)= ${ax}^{2} + bx + c$ , then evaluate: $\frac{β}{\propto \propto + b} + \frac{\propto}{\propto β + b}$ _________.

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a

\frac{- 2}{a}

b

\frac{- 2}{b}

c

\frac{- 1}{a}

d

\frac{- 1}{b}

answer is A.

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Detailed Solution

As per the question, α and β are the zeros or roots of the quadratic polynomial f(x)=

{ax}^{2} + bx + c

, this is a general quadratic equation so, their roots are also general.
Therefore, as per the given equation sum and product of these roots will be as below:
As we know the sum of the root of an equation, α+β=

- \frac{b}{a}

And product of roots, αβ=

\frac{c}{a}

Now, we will simplify the equation which we want to evaluate by taking L.C.M.

\frac{β}{\propto \propto + b} + \frac{\propto}{\propto β + b}

While taking L.C.M. In numerator, we will multiply β with (aβ+b), α with (aα+b) and in denominator we will multiply (aα+b)(aα+b) with (aβ+b)
Here, the result of L.C.M.
=

\frac{a β^{2} + βb + a α^{2} + bα}{a^{2} αβ + aαb + abβ + b^{2}}

Now, we will take commons from the numerator: a common from first, third terms and take b common from second, fourth terms.
Take commons from the denominator: ab second and third terms.
After that we get =

\frac{a (β^{2} + α^{2}) + b (\propto + β)}{ab (α + β) + a^{2} αβ + b^{2}}

We have an identity

a^{2} + b^{2} = {(a + b)}^{2} - 2 ab

, which we can put in the first term of the numerator.
Then, we get =

\frac{a ({(β + α)}^{2} - 2 \propto β) + b (α + β)}{ab (α + β) + a^{2} αβ + b^{2}}

......(1)
Now, we will put the values of α+β and αβ in equation (1)
After putting values we get:

\frac{a ({(\frac{- b}{a})}^{2} - \frac{c}{a}) + b (\frac{- b}{a})}{ab (\frac{- b}{a}) + a^{2} \times \frac{c}{a} + b^{2}}

We will simplify the above equation by expanding each term.
=:

\frac{a (\frac{b^{2} - 2 ac}{a \times a}) + b (\frac{- b}{a})}{ab (\frac{- b}{a}) + \frac{a \times a \times c}{a} + b^{2}}

Here, we cancel out negative b2b2 with positive b2b2 and aa , cc present in numerator and denominator.
=

\frac{\frac{b^{2} - 2 ca - b^{2}}{a}}{- b^{2} + ac + b^{2}}

=

\frac{- 2 c}{ac}

⇒

\frac{- 2}{a}

.
Finally, this is the result for above evaluation

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