If α and β are the roots of the equation a x^2+b x+c=0 , where a ≠ 0 , then (a α+b)(a β+b) is equal to:

Correct answer: C. c a α, β are the roots of the given equation a x^2+b x+c=0 Sum of roots =α+β=-(b)/(a) and product of the roots =α × β=(c)/(a) Now, value of (a α+b)(a β+b)=a^2 α β+a b α+a b β+b^2=a^2(α β)+a b(α+β)+b^2=a^2((c)/(a))+a b(-(b)/(a))+b^2=a c-b^2+b^2=a c .

NDA 2014

Exam: NDA
Year: 2014
Subject: Mathematics
Chapter: Quadratic Equation

NDA 2014 · Mathematics · Quadratic Equation

If α and β are the roots of the equation a x^2+b x+c=0 , where a ≠...

\alpha

and

\beta

are the roots of the equation

a x^2+b x+c=0

, where

a \neq 0

, then

(a \alpha+b)(a \beta+b)

is equal to:

Options

A $a b$
B $b c$
C $c a$
D $a b c$

Correct answer

C. $c a$

Step-by-step solution

\because \alpha, \beta

are the roots of the given equation

a x^2+b x+c=0\therefore

Sum of roots

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

and product of the roots

=\alpha \times \beta=\frac{c}{a}

Now, value of

(a \alpha+b)(a \beta+b)=a^2 \alpha \beta+a b \alpha+a b \beta+b^2=a^2(\alpha \beta)+a b(\alpha+\beta)+b^2=a^2\left(\frac{c}{a}\right)+a b\left(-\frac{b}{a}\right)+b^2=a c-b^2+b^2=a c

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