Question

A value of $b$ for which the equations $x^2+bx-1=0$ and $x^2+x+b=0$ have one common root:

• A. $-\sqrt{2}$
• B. $-i\sqrt{3}$
• C. $i\sqrt{5}$
• D. $2$

Answer 1

The correct option is B $-i\sqrt{3}$

$\bullet$ Only one root is common:

Let $\alpha$ be the common root, then

$\frac{{\alpha }^2}{b_1{c}_2-b_2{c}_1}=\frac{\alpha }{a_2{c}_1-a_1{c}_2}=\frac{1}{a_1{b}_2-a_2{b}_1}$

Above Formula is used to find common root between

$a_1{x}^2+b_{1}x+c_1=0$

$a_2{x}^2+b_{2}x+c_2=0$

Now comparing given equations with above mentioned standard Equation

we get

$a_1=1,b_1=b,c_1=-1$ and $a_2=1,b_2=1,c_2=b$

Now ptting above values in the given formula we get

$\frac{{\alpha }^2}{b^2-(-1)}=\frac{\alpha }{-1-b}=\frac{1}{1-b}$

now ${\alpha }^2=\frac{b^2+1}{1-b}$ ...... (1)

and $\alpha =\frac{b+1}{b-1}$ ....... (2)

Putting value of $\alpha$ in (1), we get

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

Solving further we get

$b^2+2b+1=b^2-b^3+1-b$

$\Rightarrow$ $3b=-b^3$

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

$\Rightarrow$ $b=+i\sqrt{3},-i\sqrt{3}$

Therefore answer is $\left(B\right)$