If α and β are the roots of a x²+b x+c=0 then the equation a x²-b x(x-1)+c(x-1)²= 0 has roots

Correct Answer is : (α1+α),(β1+β)≤ft(α/1+α),≤ft(β/1+β)(1+αα),(1+ββ)

Correct Answer is : (α1+α),(β1+β)≤ft(α/1+α),≤ft(β/1+β)(1+αα),(1+ββ)

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BITSAT 2011 Paper

Section: Mathematics

Question : 86 of 150

Marks: +1, -0

\alpha

\beta

a x^{2}+b x+c=0

a x^{2}-b x(x-1)+c(x-1)^{2}=

0 has roots

Solution:

Since

\alpha

and

\beta

are the roots of

a x^{2}+b x+c=0

therefore

\alpha+\beta=-\frac{b}{a}, \alpha\beta=\frac{c}{a}

The equation

a x^{2}-b x(x-1)+c(x-1)^{2}=0

can be written as

x^{2}(a-b+c)+ (b-2 c) x+c=0

Sum of the roots of this equation is

S=-\frac{b-2c}{a-b+c}=\frac{-b+2c}{a-b+x}=\frac{-\frac{b}{a}+\frac{2c}{a}}{1-\frac{b}{a}+\frac{c}{a}}

\Rightarrow S=\frac{\alpha+\beta+2\alpha\beta}{1+\alpha+\beta+\alpha\beta}=\frac{\alpha}{\alpha+1}+\frac{\beta}{\beta+1}

Product of the roots

=\frac{c}{a-b+c}

\Rightarrow P=\frac{\frac{c}{a}}{1-\frac{b}{a}+\frac{c}{a}}

\Rightarrow P=\frac{\alpha\beta}{1+\alpha+\beta+\alpha\beta}=\frac{\alpha}{a+1}\cdot\frac{\beta}{\beta+1}

Thus,

a x^{2}-b x(x-1)+c(x-1)^{2}=0

has

\frac{\alpha}{a+1}, \frac{\beta}{\beta+1}

as its two roots.

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