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BITSAT 2015
Exam
BITSAT
Year
2015
Subject
Mathematics
Chapter
Complex Number

BITSAT 2015 · Mathematics · Complex Number

The root of the equation 2(1+i) x^ 2 -4(2-i) x-5-3 i=0 which has...

The root of the equation 2(1+i)x2−4(2−i)x−5−3i=02(1+i) x^{2}-4(2-i) x-5-3 i=0 which has greater modulus is

Options

  1. A3−5i2\frac{3-5 i}{2}
  2. B5−3i2\frac{5-3 \mathrm{i}}{2}
  3. C3−i2\frac{3-i}{2}
  4. Dnone

Correct answer

A. 3−5i2\frac{3-5 i}{2}

Step-by-step solution

 Roots =4(2−i)±16(2−i)2+8(1+i)(5+3i)4(1+i)=4−i1+i or −i1+i=3−5i2 or −1−i2\begin{aligned} \text{ Roots } &=\frac{4(2-i) \pm \sqrt{16(2-i)^{2}+8(1+i)(5+3 i)}}{4(1+i)} \\ &=\frac{4-i}{1+i} \text{ or } \frac{-i}{1+i}=\frac{3-5 i}{2} \text{ or } \frac{-1-i}{2} \end{aligned}
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