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NDA 2011
Exam
NDA
Year
2011
Subject
Mathematics
Chapter
Quadratic Equation

NDA 2011 · Mathematics · Quadratic Equation

Let α, β be the roots of the equation (x-a)(x-b)=c, c ≠ 0 . Then...

Let α,β\alpha, \beta be the roots of the equation (x−a)(x−b)=c,c≠0(x-a)(x-b)=c, c \neq 0. Then the roots of the equation (x−α)(x−β)+c=0(x-\alpha)(x-\beta)+c=0 are

Options

  1. Aa,ca, c
  2. Bb,cb, c
  3. Ca,ba, b
  4. Da+b,a+ca+b, a+c

Correct answer

C. a,ba, b

Step-by-step solution

The given equation (x−a)(x−b)=c(x-a)(x-b)=c can be written as x2−(a+b)x+(ab−c)=0x^2-(a+b) x+(a b-c)=0 ...(i) ∵α,β\because \alpha, \beta are the roots of the equation \left.∴α+β=a+bαβ=ab−c\begin{array}{rl}\therefore \quad \alpha+\beta & =a+b \\ \alpha \beta & =a b-c\end{array} ...(ii)IfλandIf \lambda and\deltaaretherootsofequationare the roots of equation(x−α)(x−β)+c=0x2−(α+β)x+(αβ+c)=0∴λ+δ=α+β=a+b\begin{aligned}(x-\alpha)(x-\beta)+c & =0 \\ x^2-(\alpha+\beta) x+(\alpha \beta+c) & =0 \\ \therefore \quad \lambda+\delta=\alpha+\beta & =a+b\end{aligned}andλ.δ=αβ+c=aband \quad \lambda . \delta=\alpha \beta+c=a b by …(ii)\ldots(i i) Clearly, the roots λandδ\lambda and \delta are aandba and b.
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