Verified Solution • 2025–26 Session moderate

QJEEM25Maf18b: JEE Main 2025 Mathematics - Class 11 JEEM25Maf18b Solution

If α+iβ\alpha + i\beta and γ+iδ\gamma + i\delta are the roots of x2−(3−2i)x−(2i−2)=0x^2 - (3 - 2i)x - (2i - 2) = 0 , i=−1i = \sqrt{-1} , then αγ+βδ\alpha \gamma + \beta \delta is equal to:

Multiple Choice Options

  • A 2
  • B -6
  • C 6
  • D -2

Step-by-Step Verified Solution

amp;x2−(3−2i)x−(2i−2)=0amp;x=(3−2i)±(3−2i)2−4(1)(−(2i−2))2(1)amp;==(3−2i)±9−4−12i+8i−82amp;==3−2i±−3−4i2amp;=3−2i±(1)2+(2i)2−2(1)(2i)2amp;=3−2i±(1−2i)2amp;⇒3−2i+1−2i2 or 3−2i−1+2i2amp;⇒2−2i or 1+0iamp; So αγ+βδ=2(1)+(−2)(0)=2\begin{aligned}& x^2-(3-2 i) x-(2 i-2)=0 \\& x=\frac{(3-2 i) \pm \sqrt{(3-2 i)^2-4(1)(-(2 i-2))}}{2(1)} \\& ==\frac{(3-2 i) \pm \sqrt{9-4-12 i+8 i-8}}{2} \\& ==\frac{3-2 i \pm \sqrt{-3-4 i}}{2} \\& =\frac{3-2 i \pm \sqrt{(1)^2+(2 i)^2-2(1)(2 i)}}{2} \\& =\frac{3-2 \mathrm{i} \pm(1-2 \mathrm{i})}{2} \\& \Rightarrow \frac{3-2 \mathrm{i}+1-2 \mathrm{i}}{2} \text { or } \frac{3-2 \mathrm{i}-1+2 \mathrm{i}}{2} \\& \Rightarrow 2-2 \mathrm{i} \text { or } 1+0 \mathrm{i} \\& \text { So } \alpha \gamma+\beta \delta=2(1)+(-2)(0)=2\end{aligned}

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