discriminant?

Hi all, I thought this question would be quite easy, but I'm having trouble with it, can anyone help? Rep to anyone who can
--
Find the range of values of a for which
(2-3a)x^2 + (4-a)x + 2
has no real roots
--
I thought it would be simply where b^2 - 4ac < 0
but I end up with a^2 < 16
so a < +- 4
or just a < 4

but the back of the book says -16<a<0
:confused:

Reply 1

Zaq

b^2 - 4ac < 0
16 + a^2 - 8a - 16 + 24a< 0
a^2 + 16a < 0
a ( a +16 ) < 0

a < 0
and
a< -16

range of a : { -16 < a < 0 }
:wink:

Reply 2

kimoni

aah ok, i am a nilly

one thing though

Zaq

a< -16

range of a : { -16 < a < 0 }
:wink:

if a is smaller than -16, how does that change to being greater to -16?

Reply 3

Zaq

draw the line diagram.

Reply 4

terryb

a < 0
and
a < -16

Not quite true. Of course, if both parts are negative then the product is positive.

range of a : { -16 < a < 0 }

This is right, though. For the product to be negative, the roots have to be of different signs, which is what this says.

Terry

Reply 5

Zaq

shud i write 'or' rather than an 'an' ?

ya..a^2 = + ve
but isn't tat a^2 + 16a < 0 ?
v still get < 0 if v use the range

Reply 6

terryb

What we're after is something along the lines of:

a < 0

AND

a > -16

because a has to be between the roots of the corresponding equation.
(It's obvious you know what you mean, just a typo)

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