IIT JEE , AIEEE Forum - Mathematics , Algebra - theory of eqn---challenge

ravitej

1.if the eqn x⁴-2x²+k=0 has 4 real roots then interval for k is ______ans:[0,4]

2.ax³+bx²+cx+d=0 has one real and 2 imaginary roots for real values of a,b,c,d

where a!=0,b!=0 then which of the following is most appropriate ans

1. b^2<=3ac

2.b^2<3ac

3. b^2>3ac ans:1

amaron

For the first one,I think the answer is k belongs to [0,1]
as,

let x² =y,then

y² -2y +k=0

for the original eqn.to have 4 real roots, the above eqn must have both real roots and both positive roots as x² =y

therefore D>=0 implies k<=1

and k>=0.
Please check.

Moderator

Dear Ravitej,

Experts only answer one question at a time, please let us know which question needs to be answered first and post other queries again.

~ moderator

ravitej

i also thoughtso.. in maths today march2006 practice paper ans is given so......

i want ans for 2nd que at first moderator........

ravitej

any body...............
its urgent

shailendra

hi ravi !!!!

fist chek the ans of d fist question.

i thnk it should be[0,1]

iitkgp_bipin

Let f(x) = ax³ + bx² + cx + d

f'(x) = 3ax² + 2bx + c

It is possible that f(x) has only one real root when it is increasing or decreasing.
Then discriminant of f'(x) should be less than zero.
D = (2b)² - 4(3a)(c) < 0

b² < 3ac

rajat

sir,i did not understand ur solution . plzz. re-explain

iitkgp_bipin

See if the function is only increasing or only decreasing then f'(x) is either negative or positive throughout and hence discriminant is always less than zero.

Hope u understood now.

rajat

ya,got it .thanx

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