Polynomial Question help please (1 Viewer)

[bored.of.u]

i need help with this question thanx in advance sory for my failing...

two roots of x^3 + mx^2 + 15x - 7 = 0 are equal and rational. Find m.

ok ive done this so far

let the roots of the equation be a, a and b.

2a + b = -m
a^2 + 2ab = 15
a^2b = 7

so when i rearrange the first equation 2a + b = -m

b = -m-2a i substitue this into the second equation a^2 + 2ab = 15

a^2+2a(-m-2a) = 15
a^2-2ma - 4a^2 -15 =0
a^2-(2m+4a)a-15=0

where do i go from here or have i made a mistake in my earlier steps??

<O</O

 

[jpmeijer]

let the roots of the equation be a, a and b.

2a + b = -m ...(1)
a^2 + 2ab = 15 ...(2)
a^2b = 7 ...(3)

rearrange (3) giving b = 7/a²

sub into (2) giving:
a² + 2a*(7/a²) = 15
a³ - 15a + 14 = 0
(a - 1)(a² + a - 14) = 0 (found the root a = 1 and then factored by inspection... don't know if this is 2unit?)
Therefore a = 1 (disregarding the solutions to the quadratic because they are irrational)

And then it's easy to find b = 7 and m = -9

 

[bored.of.u]

rearrange (3) giving b = 7/a²

sub into (2) giving:
a² + 2a*(7/a²) = 15
a³ - 15a + 14 = 0
(a - 1)(a² + a - 14) = 0 (found the root a = 1 and then factored by inspection... don't know if this is 2unit?)
Therefore a = 1 (disregarding the solutions to the quadratic because they are irrational)

And then it's easy to find b = 7 and m = -9

wow yeah how could i be so stupid...>< thanx jpmeijer