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Root

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Root

by cartera » Thu Jan 22, 2009 8:16 am
If r and s are root of x^2+bx+c=0, rs<0?
1). b<0
2). c<0

OA B

I dont catch it
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by techwiz » Thu Jan 22, 2009 1:21 pm
To ensure that rs <0, the r and s must be of opposite sign. Which means b^2 - 4c > b^2 [ as per Quadratic equation solution]

Which means c < 0. Hece statement 2 is alone sufficient.
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by welcome » Thu Jan 22, 2009 1:32 pm
Solution of x^2+bX+c=0 will be

(-b+(b^2-4c)^1/2)/2 = r
(-b-(b^2-4c)^1/2)/2 = s

so r*s = b^2-(b^2-4c)/4
=> rs = 4c/4 = c

means rs is equal to C.

1) value of b will not affect value of rs - NOT SUff
2) value of C will direclt effect rs - Answer

Answerd : B.
Shubham.
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by techwiz » Thu Jan 22, 2009 2:23 pm
To ensure that rs <0, the r and s must be of opposite sign. Which means b^2 - 4c > b^2 [ as per Quadratic equation solution]

Which means c < 0. Hece statement 2 is alone sufficient.
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Re: Root

by aroon7 » Sat Jan 24, 2009 9:50 am
cartera wrote:If r and s are root of x^2+bx+c=0, rs<0?
1). b<0
2). c<0

OA B

I dont catch it
r*s is product of the roots of this quadratic equation
ie constant term / coeff of x^2
= c
if c is less than 0, then rs < 0
--------------------------
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by GID09 » Thu Jan 29, 2009 1:56 pm
If r and S are the roots of the quadratic equation x^2+bX+C = 0 ...(1)
then (X-r)(X-s)=0
Upon simplification
X^2 -(r+s)x+rs = 0 ....(2)

Comparing (2) with (1)
b= -r-s
c=rs

Case 1
b<0 => -r-s<0 => Implies nothing ...insuff

Case 2
c<0 => rs< 0 => suff

Hence B
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