JEE MAIN - Mathematics (2016 - 9th April Morning Slot - No. 5)
If the equations x2 + bx−1 = 0 and x2 + x + b = 0 have a common root different from −1, then $$\left| b \right|$$ is equal to :
$$\sqrt 2 $$
2
3
$$\sqrt 3 $$
Explanation
Given,
x2 + bx $$-$$ 1 = 0 . . . . .(1)
and x2 + x + b = 0 . . . . . (2)
Performing (1) $$-$$ (2) we get,
bx $$-$$ 1 $$-$$ x $$-$$ b = 0
$$ \Rightarrow $$ x(b $$-$$ 1) = b + 1
$$ \Rightarrow $$ x = $${{b + 1} \over {b - 1}}$$
putting value of x in equation (2),
$${\left( {{{b + 1} \over {b - 1}}} \right)^2} + \left( {{{b + 1} \over {b - 1}}} \right) + b = 0$$
$$ \Rightarrow $$ (b + 1)2 + (b + 1) (b $$-$$ 1) + b (b $$-$$ 1)2 = 0
$$ \Rightarrow $$ b2 + 2b + 1 + b2 $$-$$ 1 + b (b2 $$-$$ 2b + 1) = 0
$$ \Rightarrow $$ 2b3 + 2b + b3 $$-$$ 2b2 + b = 0
$$ \Rightarrow $$ b3 + 3b = 0
$$ \Rightarrow $$ b(b2 + 3) = 0
b2 = $$-$$ 3, b = 0
$$ \therefore $$ b = $$ \pm \sqrt 3 i$$
$$ \Rightarrow $$ $$\left| b \right|$$ = $$\sqrt 3 $$
x2 + bx $$-$$ 1 = 0 . . . . .(1)
and x2 + x + b = 0 . . . . . (2)
Performing (1) $$-$$ (2) we get,
bx $$-$$ 1 $$-$$ x $$-$$ b = 0
$$ \Rightarrow $$ x(b $$-$$ 1) = b + 1
$$ \Rightarrow $$ x = $${{b + 1} \over {b - 1}}$$
putting value of x in equation (2),
$${\left( {{{b + 1} \over {b - 1}}} \right)^2} + \left( {{{b + 1} \over {b - 1}}} \right) + b = 0$$
$$ \Rightarrow $$ (b + 1)2 + (b + 1) (b $$-$$ 1) + b (b $$-$$ 1)2 = 0
$$ \Rightarrow $$ b2 + 2b + 1 + b2 $$-$$ 1 + b (b2 $$-$$ 2b + 1) = 0
$$ \Rightarrow $$ 2b3 + 2b + b3 $$-$$ 2b2 + b = 0
$$ \Rightarrow $$ b3 + 3b = 0
$$ \Rightarrow $$ b(b2 + 3) = 0
b2 = $$-$$ 3, b = 0
$$ \therefore $$ b = $$ \pm \sqrt 3 i$$
$$ \Rightarrow $$ $$\left| b \right|$$ = $$\sqrt 3 $$
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