Discussion Forum

The value of  b for which the equations $x^{2}+bx-1=0,x^{2}+x+b=0$ have  one root in common  is 

Answer:

Explanation:

If $a_{1}x^{2}+b_{1}x+c_{1}=0$

 and $a_{2}x^{2}+b_{2}x+c_{2}=0$

 Have  a comon  real root , then 

$\Rightarrow$  $(a_{1}c_{2}-a_{2}c_{1})^{2}= (b_{1}c_{2}-b_{2}c_{1})(a_{1}b_{2}-a_{2}b_{1})$

 $\because$    $x^{2}+bx-1=0$, 

                      $x^{2}+x+b$   , have a common root

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

$\Rightarrow$  $b^{2}+2b+1=b^{2}-b^{3}+1-b$

$\Rightarrow$ $b^{3}+3b=0\Rightarrow b(b^{2}+3)=0$

$\Rightarrow$  $b=0, \pm \sqrt{3} i$