Mathematics - Quadratic Equation Question with Solution | TestHub

Let common root be$x$
$x^2+bx-1=0$...$\left(1\right)$

$x^2+x+b=0$...$\left(2\right)$

$\left(1\right)-\left(2\right)$, we get$x=\frac{b+1}{b-1}$

Put$x$in Equation$\left(1\right)$, we get

${\left(\frac{b+1}{b-1}\right)}^2+\left(\frac{b+1}{b-1}\right)+b=0$

$\Rightarrow$${\left(b+1\right)}^2 + \left(b+1\right) \left(b-1\right)+b{\left(b-1\right)}^2=0$

$\Rightarrow$$b^2+1+2b+b^2-1+b\left(b^2-2b+1\right)=0$

$\Rightarrow$$2b^2+2b+b^3-2b^2+b=0$

$\Rightarrow$$b^3+3b=0$
$\Rightarrow$$b\left(b^2+3\right)=0\Rightarrow b=0$or$b^2=-3$

When$b=0$then common root$x=-1$
But the given common root$x\ne -1$

$\therefore$$b^2= -3$

$b= \pm \sqrt{3}i$

$\left|b\right|= \sqrt{3}$