biquadratic equation

juantheron · 2011-10-25 23:21:26

Find the values of

for which the equation

has atleast one real root.

bobbym · 2011-10-26 00:07:39

With the subs of v = x^2+x+2 you get:

Substituting back.

Solve using the quadratic formula.

Set the discriminant equal to 0.

a=5 is indeterminate.

Trying a = 19 / 3 your equation has 1 real root of - 1 / 2 with multiplicity 4.

Since there are more answers you need to solve the inequality:

You should get:

I have found no solutions outside of 5 < a <= 19 / 3

juantheron · 2011-10-26 01:20:55

juantheron wrote:

Find the values of
for which the equation
has atleast one real root.

Thanks bobbym for nice solution.

I have solved in this way

Now here

so We can divide by

now Let

where

So

for at least one real roots.

so

Now from here what can i do..

help required.

Last edited by juantheron (2011-10-26 01:43:55)

bobbym · 2011-10-26 01:28:22

Hi;

I am not sure. It seems that for your inequality a can be anything.

I do not agree with this:

juantheron · 2011-10-26 01:42:24

oh sorry actually it is

bobbym · 2011-10-26 01:51:42

Hi;

I think that your final answer is the problem.

a = 2 works.
a= -1 works
a = 100 works
a = -√123456 works ...

I think it is better to just reduce the original problem to a quadratic as I have done.

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