If the equation x^{2}+bx+45=0 (b in R) has co | vedclass

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(D) Let the roots of the equation $x^{2}+bx+45=0$ be $z$ and $\bar{z}$.Since the roots are complex,the discriminant $D < 0$,so $b^{2}-4(45) < 0$,which

Vedclass · Accepted Answer

$b^{2}-b=30$

Vedclass · Answer

(D) Let the roots of the equation $x^{2}+bx+45=0$ be $z$ and $\bar{z}$.Since the roots are complex,the discriminant $D The roots are $z = \frac{-b \pm i\sqrt{180-b^{2}}}{2}$.Given $|z+1| = 2\sqrt{10}$,we have $|z+1|^{2} = 40$.Let $z = x+iy$,then $x = -b/2$ and $y = \pm \frac{\sqrt{180-b^{2}}}{2}$.So,$(x+1)^{2} + y^{2} = 40$.Substituting the values,$(1 - b/2)^{2} + \frac{180-b^{2}}{4} = 40$.$1 - b + \frac{b^{2}}{4} + 45 - \frac{b^{2}}{4} = 40$.$46 - b = 40$,which gives $b = 6$.Now,checking the options for $b=6$:$b^{2}-b = 36-6 = 30$.Thus,the correct option is $D$.

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

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