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Complex Numbers question

2020 · 8 Jan · Shift 1 · Q35
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Complex Numbers question

2020 · 8 Jan · Shift 1 · Q35

JEE MainMathematicsComplex NumbersMCQ+4 / −1
If the equation, x2 + bx + 45 = 0 (b ∈\in∈ R) has conjugate complex roots and they satisfy |z +1| = 2 10\sqrt {10}10​ , then :
  1. A
    b2 – b = 42
  2. B
    b2 + b = 12
  3. C
    b2 + b = 72
  4. D
    b2 – b = 30
View written solutionFree

Correct answer: D

  1. Write the roots in complex form

Since the quadratic x2+bx+45=0x^2+bx+45=0x2+bx+45=0 has real coefficients and conjugate complex roots, let the roots be z=a+ic,zˉ=a−icz=a+i c, \quad \bar z=a-i cz=a+ic,zˉ=a−ic with c≠0c\ne 0c=0.

  1. Use sum and product of roots

For the quadratic x2+bx+45=0x^2+bx+45=0x2+bx+45=0,

  • sum of roots =−b= -b=−b
  • product of roots =45=45=45

So, z+zˉ=2a=−b  ⟹  a=−b2z+\bar z=2a=-b \implies a=-\frac b2z+zˉ=2a=−b⟹a=−2b​

and zzˉ=∣z∣2=a2+c2=45.z\bar z=|z|^2=a^2+c^2=45.zzˉ=∣z∣2=a2+c2=45.

  1. Use the given modulus condition

Given: ∣z+1∣=210|z+1|=2\sqrt{10}∣z+1∣=210​ Squaring both sides, ∣z+1∣2=40.|z+1|^2=40.∣z+1∣2=40.

If z=a+icz=a+icz=a+ic, then ∣z+1∣2=(a+1)2+c2=40.|z+1|^2=(a+1)^2+c^2=40.∣z+1∣2=(a+1)2+c2=40.

But from a2+c2=45a^2+c^2=45a2+c2=45, substitute: (a+1)2+c2=a2+2a+1+c2=45+2a+1=46+2a.(a+1)^2+c^2=a^2+2a+1+c^2=45+2a+1=46+2a.(a+1)2+c2=a2+2a+1+c2=45+2a+1=46+2a. Thus, 46+2a=4046+2a=4046+2a=40 2a=−62a=-62a=−6 a=−3.a=-3.a=−3.

  1. Find bbb

Since a=−b2,a=-\frac b2,a=−2b​, we get −3=−b2  ⟹  b=6.-3=-\frac b2 \implies b=6.−3=−2b​⟹b=6.

  1. Check the options

Now compute:

  • A: b2−b=36−6=30≠42b^2-b=36-6=30 \ne 42b2−b=36−6=30=42
  • B: b2+b=36+6=42≠12b^2+b=36+6=42 \ne 12b2+b=36+6=42=12
  • C: b2+b=42≠72b^2+b=42 \ne 72b2+b=42=72
  • D: b2−b=36−6=30b^2-b=36-6=30b2−b=36−6=30

So the correct option is D.\boxed{D}.D​.

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