Brainiac 7

Algebra Level 2

A value for ' $b$ ' such that

$x^{2}$ + $bx$ $-1$ =0

$x^{2} +x+b=0$

have a common root!

source;IIT-paper 2

isqrt5 sqrt2 -sqrt2 -isqrt(3)

2 solutions

Ahmed Arup Shihab
Feb 17, 2015

Subtracting second equation from the first,

$x(b-1)-(1+b)=0$

$\Rightarrow x=\frac{b+1}{b-1}$ , which is the common root.

Substitute this value of $x$ in either first or second equation we get,

$b=\pm i\sqrt{3}$

Anshuman Padhi
Sep 16, 2014

The two soln. are x^2 +bx-1=0 and x^2+x+b=0. by condition of common root, let a be the common root, a^2/(b^2+1)=a/(-1-b)=1/(1-b). thus, solving for b, we get b= -isqrt(3)

Sir, I think I've a better solution:

Let\quad \alpha \quad be\quad the\quad common\quad root\quad of\quad the\quad two\quad equations.\ Hence\quad { \alpha }^{ 2 }+\alpha b-1=0\quad and\quad { \alpha }^{ 2 }+{ \alpha }+b=0.\ Solving\quad the\quad two\quad equations:\ \frac { { \alpha }^{ 2 } }{ b^{ 2 }+1 } =\frac { { \alpha } }{ -(1+b) } =\frac { 1 }{ 1-b } \ \Longrightarrow { \quad { -(1+b)} }^{ 2 }=({ b }^{ 2 }+1)(1-b)\ \Longrightarrow { \quad (1+b) }^{ 2 }=({ b }^{ 2 }+1)(1-b)\ \Longrightarrow \quad -3={ b }^{ 2 }\ \Longrightarrow \quad b=\sqrt { 3i }

Neeraj Snappy - 6 years, 6 months ago

Brainiac 7

2 solutions

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