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NDA 2017
Exam
NDA
Year
2017
Subject
Mathematics
Chapter
Quadratic Equation

NDA 2017 · Mathematics · Quadratic Equation

The roots of the equation (q-r) x^2+(r-p) x+(p-q)=0 are

The roots of the equation (q−r)x2+(r−p)x+(p−q)=0(q-r) x^2+(r-p) x+(p-q)=0 are

Options

  1. A(r−p)/(q−r),1/2(\mathrm{r}-\mathrm{p}) /(\mathrm{q}-\mathrm{r}), 1 / 2
  2. B(p−q)/(q−r),1(p-q) /(q-r), 1
  3. C(q−r)/(p−q),1(\mathrm{q}-\mathrm{r}) /(\mathrm{p}-\mathrm{q}), 1
  4. D(r−p)/(p−q),1/2(\mathrm{r}-\mathrm{p}) /(\mathrm{p}-\mathrm{q}), 1 / 2

Correct answer

B. (p−q)/(q−r),1(p-q) /(q-r), 1

Step-by-step solution

Given equation, (q−r)x2+(r−p)x+(p−q)=0(q-r) x^2+(r-p) x+(p-q)=0 On observing the equation, it is clear that 1 is root of equation. If x=1x=1, then q−r+r−p+p−q=0q-r+r-p+p-q=0. ∴1\therefore 1 is one root of given equation. Since, the given equation is quadratic equation, we know that product of roots is ca\frac{\mathrm{c}}{\mathrm{a}}. Let the second root be α\alpha. ∴(1)(α)=p−qq−r⇒α=p−qq−r\therefore \quad(1)(\alpha)=\frac{p-q}{q-r} \Rightarrow \alpha=\frac{p-q}{q-r}.
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