Q33 / 90 · JEE Main 2016 (09 Apr Online) · Mathematics

If the equations $x^{2} + b x - 1 = 0$ and $x^{2} + x + b = 0$ have a common root different from $- 1,$ then $|b|$ is equal to :

A
$2$
B
$3$
Correct answer
C
$\sqrt{3}$
D
$\sqrt{2}$

Solution

Let common root be

x

x^{2} + b x - 1 = 0

...

(1)

x^{2} + x + b = 0

...

(2)

(1) - (2)

, we get

x = \frac{b + 1}{b - 1}

Put

x

in Equation

(1)

, we get

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

\Rightarrow

{(b + 1)}^{2} + (b + 1) (b - 1) + b {(b - 1)}^{2} = 0

\Rightarrow

b^{2} + 1 + 2 b + b^{2} - 1 + b (b^{2} - 2 b + 1) = 0

\Rightarrow

2 b^{2} + 2 b + b^{3} - 2 b^{2} + b = 0

\Rightarrow

b^{3} + 3 b = 0

\Rightarrow

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

or

b^{2} = - 3

When

b = 0

then common root

x = - 1

But the given common root

x \neq - 1

∴

b^{2} = - 3

b = \pm \sqrt{3} i

|b| = \sqrt{3}

Hence, the correct answer is option: 'C'.

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