Question

The value of b for which the equations x${}^2$ +bx -1=0 and 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

Answer: (B)

-i $\sqrt{3}$

x${}^2$+x+b=0, x${}^2$+bx-1 = 0
$\Rightarrow$ = $\frac{x^2}{b^2+1}$ = $\frac{x}{-1-b}$= $\frac{1}{1-b}$
$\Rightarrow$ = x= $\frac{(b^{[}2}{+}1\left)\right]-(b+1)$= $\frac{-(b+1)}{(1-b)}$
$\to$ (b${}^2$+1)(1-b)=(b +1)${}^2$
$\to$ b${}^2$ -b${}^3$ + 1 -b = b${}^2$ +2b + 1
$\Rightarrow$ b${}^3$+3b=0 $\Rightarrow$ b=0,b${}^2$=-3
$\Rightarrow$ b = 0, $\pm$ $\sqrt{3}$i
First let us suppose that all four roots are real and distined
Let f(x) = x${}^4$ - 4x${}^3$ + 12x${}^2$ + x -1 = 0
Let f(x) = x${}^4$ - 4x${}^3$ + 12x${}^2$ + x -1 = 0
$\Rightarrow$ f'(x) = 4x${}^3$ - 12x${}^2$ + 24x+1
Must have three distinct real roots
f"(x)=0 $\Rightarrow$ 12x${}^2$ -24x + 24 = 0
Most have two distinct real roots which is a contradiztion because 12x${}^2$ - 24x + 24 = 0
or x${}^2$ - 2x + 2 = Q
$\Rightarrow$ D < 0
Hence f(x) = 0 can not have all four equal roots.As f(0) = -1, f(1) = 9, f(-1) = 15
$\Rightarrow$ f(x) = 0 must have two distinct real roots are in [ -1,0 [ and other in [0,1[.