• Oct 26th 2010, 12:10 AM
prasum
if exactly one root of equation x^2-(k-1)x+k(k+4)=0 lies between the roots of equation x^2-(k+3)x+k+2=0 then value of k=

i have solved this graphically

can yu solve this mathematically
• Oct 26th 2010, 12:44 AM
earboth
Quote:

Originally Posted by prasum
if exactly one root of equation x^2-(k-1)x+k(k+4)=0 lies between the roots of equation x^2-(k+3)x+k+2=0 then value of k=

i have solved this graphically

can yu solve this mathematically

1. According to the theorem of Viéte (or do you call him Vieta?) the mean of the 2 solutions of the 2nd equation is $\dfrac{k+3}2$

2. Solve the 1rt equation using the quadratic formula. After moving some stuff around you'll get:

$x = \dfrac{(k-1)\pm\sqrt{(k-1)^2-4 \cdot 1 \cdot (k^2+4k)}}2$

3. Using the result from 1. you have to solve for k:

$(k-1)\pm \sqrt{-3k^2-18k+1}= (k+3)$

I'll leave this for you.

Spoiler:
I've got k = -1 or k = -5
• Oct 26th 2010, 04:12 AM
prasum
Quote:

Originally Posted by earboth
1. According to the theorem of Viéte (or do you call him Vieta?) the mean of the 2 solutions of the 2nd equation is $\dfrac{k+3}2$

2. Solve the 1rt equation using the quadratic formula. After moving some stuff around you'll get:

$x = \dfrac{(k-1)\pm\sqrt{(k-1)^2-4 \cdot 1 \cdot (k^2+4k)}}2$

3. Using the result from 1. you have to solve for k:

$(k-1)\pm \sqrt{-3k^2-18k+1}= (k+3)$

I'll leave this for you.

Spoiler:
I've got k = -1 or k = -5

thanks