Question

A value for b for which the equations
${x^2} + bx - 1 = 0$
${x^2} + x + b = 0$,
Have one root in common is-
A.$- \sqrt 2$
B. $i\sqrt 3$
C. $i\sqrt 5$
D. $\sqrt 2$

Answer 1

We assume the common root of both the equations as a variable and use that value to form equations with roots substituted in them.

Complete step by step answer:
We are given two quadratic equations:
${x^2} + bx - 1 = 0$ and ${x^2} + x + b = 0$ … (1)
Let both equations have a common root ‘y’.
Since ‘y’ is the root of the equations, then it must satisfy the quadratic equations.
$\Rightarrow {y^2} + by - 1 = 0$ and ${y^2} + y + b = 0$ … (2)
Now we solve for the value of common root
$\dfrac{{{y^2}}}{{{b^2} + 1}} = \dfrac{{ - y}}{{b + 1}} = \dfrac{1}{{1 - b}}$
Equate second and third fraction
$\Rightarrow \dfrac{{ - y}}{{b + 1}} = \dfrac{1}{{1 - b}}$
Cross multiply the values from RHS and LHS
$\Rightarrow y = \dfrac{{ - b - 1}}{{1 - b}}$ … (3)
Now equate first and third fraction
$\Rightarrow \dfrac{{{y^2}}}{{{b^2} + 1}} = \dfrac{1}{{1 - b}}$
Cross multiply the values from RHS and LHS
$\Rightarrow {y^2} = \dfrac{{{b^2} + 1}}{{1 - b}}$ … (4)
Since we know ${(y)^2} = {y^2}$
Substitute values of y and ${y^2}$from equations (3) and (4)
$\Rightarrow {\left( {\dfrac{{ - b - 1}}{{1 - b}}} \right)^2} = \dfrac{{{b^2} + 1}}{{1 - b}}$
Cancel same terms from denominator on both sides of the equation
$\Rightarrow \dfrac{{{{\left( { - b - 1} \right)}^2}}}{{1 - b}} = {b^2} + 1$
Use identity ${(x + y)^2} = {x^2} + {y^2} + 2xy$ to solve LHS
$\Rightarrow \dfrac{{1 + {b^2} + 2b}}{{1 - b}} = {b^2} + 1$
Cross multiply both sides of the equation
$\Rightarrow 1 + {b^2} + 2b = {b^2} + 1 - {b^3} - b$
Cancel same terms from both sides of the equation
$\Rightarrow 2b = - {b^3} - b$
Bring all terms to LHS of the equation
$\Rightarrow 2b + {b^3} + b = 0$
Add terms with same powers
$\Rightarrow {b^3} + 3b = 0$
Take b common from all terms in LHS
$\Rightarrow b({b^2} + 3) = 0$
Equate both factors to 0
$\Rightarrow {b^2} + 3 = 0$ and $b = 0$
Shift constant to RHS
$\Rightarrow {b^2} = - 3$ and $b = 0$
Take square root on both sides
$\Rightarrow b = \sqrt { - 3}$ and $b = 0$
Since $i = \sqrt { - 1}$
$\Rightarrow b = i\sqrt 3$ and $b = 0$
Since the only matching option is $b = i\sqrt 3$, the common root is $b = i\sqrt 3$.

$\therefore$Option B is correct.

Note: Many students make mistake of solving for the value of y from two formed equations by calculating root through determinant method, keep in mind this is a complex method and here we will get very confusing values for two equations and we will have check 4 pairs of values by equating each one separately which will take long.