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BITSAT 2023Paper 1
Exam
BITSAT
Year
2023
Paper
Paper 1
Subject
Mathematics
Chapter
Quadratic Equation

BITSAT 2023 · Paper 1 · Mathematics · Quadratic Equation

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

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

Options

  1. Ap−qr−p,1\frac{\mathrm{p}-\mathrm{q}}{\mathrm{r}-\mathrm{p}}, 1
  2. Bq−rp−q,1\frac{\mathrm{q}-\mathrm{r}}{\mathrm{p}-\mathrm{q}}, 1
  3. Cr−pp−q,1\frac{\mathrm{r}-\mathrm{p}}{\mathrm{p}-\mathrm{q}}, 1
  4. DNone of these

Correct answer

C. r−pp−q,1\frac{\mathrm{r}-\mathrm{p}}{\mathrm{p}-\mathrm{q}}, 1

Step-by-step solution

Given equation is
(p−q)x2+(q−r)x+(r−p)=0(p-q) x^2+(q-r) x+(r-p)=0
By using formula for finding the root viz: −b±b2−4ac2a\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}, we get
x=(r−q)±(q−r)2−4(r−p)(p−q)2(p−q)x=\frac{(r-q) \pm \sqrt{(q-r)^2-4(r-p)(p-q)}}{2(p-q)}
⇒x=(r−q)±(q+r−2p)2(p−q)=r−pp−q,1\Rightarrow \quad x=\frac{(r-q) \pm(q+r-2 p)}{2(p-q)}=\frac{r-p}{p-q}, 1
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