TRUE OR FALSE; Has $f(x) = 6acx³ + 4bcx² + 9adx + 6bd$ two opposite zeros if $c xxd > 0$ ? Thank you!

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Answer 1 · 2017-11-09T01:49:06

See below.

$6acx³ + 4bcx² + 9adx + 6bd = 0 rArr x^3+(4b)/(6a) x^2+(9d)/(6c) x + (bd)/(ac) = 0$ or

$x^3+(2b)/(3a) x^2+(3d)/(2c) x + (bd)/(ac) = 0$

Now if two roots have opposite signs by Vieta's formula

${(-(x_1-x_1+x_3) = (2b)/(3a)),(-x_1^2+x_1 x_3 - x_1 x_3 =(3d)/(2c) ),(-(-x_1^2 x_3) = (bd)/(ac) ):}$

or

${(x_3 = -(2b)/(3a)),(x_1^2=-(3d)/(2c) ),(x_1^2 x_3 =(bd)/(ac) ):}$

or concluding

$d < 0, c < 0 rArr dc > 0$

TRUE OR FALSE; Has f(x) = 6acx³ + 4bcx² + 9adx + 6bd f(x) = 6acx³ + 4bcx² + 9adx + 6bd two opposite zeros if c xxd > 0 c xxd > 0 ? Thank you!