7x-5-4x-4-9x-3-12x-2-5x-9-0-How-many-roots-of-this-equation-are-Negative-

Question Number 13438 by Nayon last updated on 20/May/17

7 x^{5} - 4 x^{4} + 9 x^{3} + 12 x^{2} + 5 x - 9 = 0

H o w m a n y r o o t s o f t h i s e q u a t i o n

a r e N e g a t i v e ?

Answered by mrW1 last updated on 20/May/17

7 x^{5} - 4 x^{4} + 9 x^{3} + 12 x^{2} + 5 x - 9 = 0

7 x^{5} - 9 x^{3} + 5 x - 4 x^{4} + 12 x^{2} - 9 = 0

x (7 x^{4} - 9 x^{2} + 5) = (4 x^{4} - 12 x^{2} + 9)

7 x (x^{4} - \frac{9}{7} x^{2} + \frac{5}{7}) = 4 (x^{4} - 3 x^{2} + \frac{9}{4})

7 x [x^{4} - 2 \times \frac{9}{14} x^{2} + {(\frac{9}{14})}^{2} + \frac{5}{7} - {(\frac{9}{14})}^{2}] = 4 [x^{4} - 2 \times \frac{3}{2} x^{2} + {(\frac{3}{2})}^{2}]

7 x [{(x^{2} - \frac{9}{14})}^{2} + \frac{59}{196}] = 4 {(x^{2} - \frac{3}{2})}^{2}

x = \frac{4 {(x^{2} - \frac{3}{2})}^{2}}{7 [{(x^{2} - \frac{9}{14})}^{2} + \frac{59}{196}]} = \frac{> 0}{> 0} > 0

\Rightarrow t h e r e i s n o n e g a t i v e r o o t!

Commented by ajfour last updated on 20/May/17

a w e s o m e!

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