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Q.

If α and β are the zeros of the quadratic polynomial f(x)=ax2+bx+c , then evaluate: β+b+β+b _________.


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a

-2a

b

-2b

c

-1a

d

-1b 

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)= ax2+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=-ba
And product of roots, αβ=ca
 Now, we will simplify the equation which we want to evaluate by taking L.C.M.
β+b+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.
=aβ2+βb+aα2+a2αβ+aαb+abβ+b2
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 =aβ2+α2+b(+β)abα+β+a2αβ+b2
We have an identity a2+b2=(a+b)2-2ab , which we can put in the first term of the numerator.
Then, we get =a((β+α)2-2β)+b(α+β)abα+β+a2αβ+b2......(1)
Now, we will put the values of α+β and αβ in equation (1)
After putting values we get: a((-ba)2-ca)+b(-ba)ab-ba+a2×ca+b2
We will simplify the above equation by expanding each term.
=: a(b2-2aca×a)+b(-ba)ab-ba+a×a×ca+b2
Here, we cancel out negative b2b2 with positive b2b2 and aa , cc present in numerator and denominator.
=b2-2ca-b2a-b2+ac+b2
=-2cac
-2a.
Finally, this is the result for above evaluation
 

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