If the equation, x2+bx+45=0,(b∈R) has conjugate complex roots and they satisfy |z+1|=210 then:

Let z=α±iβ be roots of the equation So 2α=-b and α2+β2=45 Now, |z+1|=210⇒(α+1)2+β2=40 Eliminate β, we get⇒(α+1)2-α2=-5⇒2α+1=-5⇒2α=-6 Hence b=6⇒b2-b=30

If the equation, x2+bx+45=0,(b∈R) has conjugate complex roots and they satisfy |z+1|=210 then:

Let z=α±iβ be roots of the equation So 2α=-b and α2+β2=45 Now, |z+1|=210⇒(α+1)2+β2=40 Eliminate β, we get⇒(α+1)2-α2=-5⇒2α+1=-5⇒2α=-6 Hence b=6⇒b2-b=30

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$\begin{array}{l} If the equation, x^{2} + b x + 45 = 0, (b \in R) has conjugate complex roots and they satisfy \\ | z + 1 | = 2 \sqrt{10} then: \end{array}$

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$b^{2} + b = 12$

$b^{2} + b = 72$

$b^{2} - b = 30$

$b^{2} - b = 42$

answer is B.

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Detailed Solution

$\begin{array}{l} Let z = α \pm i β be roots of the equation \\ So 2 α = - b and α^{2} + β^{2} = 45 \end{array}$

$\begin{array}{l} Now, \\ | z + 1 | = 2 \sqrt{10} \Rightarrow (α + 1)^{2} + β^{2} = 40 E l i m i n a t e β, w e g e t \\ \Rightarrow (α + 1)^{2} - α^{2} = - 5 \Rightarrow 2 α + 1 = - 5 \Rightarrow 2 α = - 6 \\ Hence b = 6 \Rightarrow b^{2} - b = 30 \end{array}$