JEE Main 2020 — Mathematics Algebra

Let $z=\alpha +i\beta$z=α+iβ be one of the roots of the equation ${x}^{2}+bx+45=0,b\in R$x2+bx+45=0,b∈R.

So, its other conjugate complex root will be $z=\bar{\alpha +i\beta }=\alpha -i\beta$z=α+iβˉ​=α−iβ.

We know that for a quadratic equation $A{x}^{2}+Bx+C=0$Ax2+Bx+C=0, the sum and product of its roots are -\frac{B}{A}&\frac{C}{A} respectively.

So, the sum of roots of the given equation, $(\alpha +i\beta )+(\alpha -i\beta )=-\frac{b}{1}\Rightarrow 2\alpha =-b......(i)$(α+iβ)+(α−iβ)=−1b​⇒2α=−b......(i).

Also, the product of roots of the given equation, $(\alpha +i\beta )(\alpha -i\beta )=\frac{45}{1}\Rightarrow {\alpha }^{2}+{\beta }^{2}=45......(\mathrm{ii})$(α+iβ)(α−iβ)=145​⇒α2+β2=45......(ii).

Now, given that $|z+1|=2\sqrt{10}$∣z+1∣=210​.

$$\Rightarrow |\alpha \pm i\beta +1|=2\sqrt{10}$$⇒∣α±iβ+1∣=210​

$$\Rightarrow \sqrt{{(\alpha +1)}^{2}+{\beta }^{2}}=2\sqrt{10}$$⇒(α+1)2+β2​=210​

$$\Rightarrow {(\alpha +1)}^{2}+{\beta }^{2}=40......(\mathrm{iii})$$⇒(α+1)2+β2=40......(iii)

Subtracting equation $(\mathrm{ii})$(ii) from equation $(\mathrm{iii})$(iii), we get

$${(\alpha +1)}^{2}-{\alpha }^{2}=-5$$(α+1)2−α2=−5

$\Rightarrow (\alpha +1+\alpha )(\alpha +1-\alpha )=-5$⇒(α+1+α)(α+1−α)=−5 $[\because {a}^{2}-{b}^{2}=(a+b)(a-b)]$[∵a2−b2=(a+b)(a−b)]

$$\Rightarrow 2\alpha +1=-5$$⇒2α+1=−5

$$\Rightarrow 2\alpha =-6$$⇒2α=−6

Comparing from equation $(i)$(i), we get

$$b=6$$b=6

$\therefore {b}^{2}-b={6}^{2}-6=36-6=30$∴b2−b=62−6=36−6=30.

Also, ${b}^{2}+b={6}^{2}+6=36+6=42$b2+b=62+6=36+6=42.

Hence, option $(a)$(a) is correct.