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The equation $x^{6}-5x^{4}+16x^{2}-72x+9=0$ has

1. exactly two distinct real roots
2. exactly three distinct real roots
3. exactly four distinct real roots
4. six different real roots
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@Tesla!, Hows it done for powers less than 6 or even lesser? All basic algebra of skul days. :).
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There is no particular answer finding roots of an polynomial is an tough problems from ages, there is no fixed way or formula for it.
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Hence by Rolle's theorem P'(x) can have at most 1 real root and  P(x) can have at most 2 real roots.

explain this pls  @ Tesla!

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@set2018 see answer by Heisenberg it is correct explanation
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I am asking how 1 real root possible with P'(x)

P(x) = $x^{6}$ − $5x^{4}$$+ 16x^{2} −72x + 9 • P(0) = 9 • P(-1)= -ve • P(4) = +ve Hence, at least 2 real roots can be clearly seen, but what about other 4 roots left ? That's why we check P^{n}\left ( x \right ), P'(x) = 6x^{5} - 20x^{3} + 32x$$-72$

$P''(x)$ $=$ $30x^{4}$ - $60x^{2}$$+32>0 for any real value of x. comparing this with a quadratic eq taking x^{2}as y we get 30y^{2}$$-60y+32$. The discriminant $\left ( b^{2}-4ac \right )$ is negative implying $P''(x)$ has no real roots

Hence by Rolle's theorem $P'(x)$ can have at most $1$ real root and  $P(x)$ can have at most $2$ real roots. Because if a function $f(x)$ has $2$ roots $x_{1}$ and $x_{2}$ then there exists a point $x ∈$ $\left [ x1,x2 \right ]$ where the curve becomes flat i.e its the root of $f'(x)$ meaning $f(x)$ has max $2$ roots

=> Exactly two distinct real roots

Option A

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but can you explain with a better approach?

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Hence by Rolle's theorem P'(x) can have at most 1 real root and  P(x) can have at most 2 real roots. Because if a function f(x) has 2 roots x1 and x2 then there exists a point x∈ [x1,x2] where the curve becomes flat i.e its the root of f′(x) meaning f(x) has max 2 roots.

Rolle's Theorem is not applicable here. Rolle's Theorem tells that if $f$ is continuous in $[a,b]$ and differentiable in $(a,b)$ and f(a)=f(b) then there exist atleast one '$c$' in $(a,b)$ for which $f'(c)=0$ Also $P''(x)>0$ ensures only that $P'(x)$ is increasing for all real $x$ but it does not imply that $P'(x)$  can have atmost or exactly one root until we prove that $\lim_{x\rightarrow -\infty }P'(x) = -\infty$ and $\lim_{x\rightarrow \infty }P'(x) = \infty$ because then it will must cross X-axis once and we already have shown that  $P''(x)>0$ means $P'(x)$ is strictly increasing.

When a function is increasing in some interval then its derivative must be +ve in that interval and when it is decreasing then its derivative must be negative. So, in the given link, f'(x) is negative in $(-\infty ,x_{0})$, So, in this interval , f(x) must be decreasing and in $(x_{0},+\infty)$, it is +ve, So, in  this interval f(x) must be increasing .

+1 vote

Using Descartes' Rule of change of sign :

No of  positive real roots of equation f(x) = 0 is given by no of sign changes in f(x) = 0 and no of negative real roots are given by no of sign changes in f(-x) = 0.

Applying this , we have :

f(x)  =   x− 5x+ 16x− 72x + 9

We can see there are 4 sign changes in the above f(x) ..So we can conclude 4 positive real roots exist..............(1)

Now

f(-x)  =  x− 5x+ 16x2  + 72x + 9

So here only 2 sign changes in the above f(-x) .So we can conclude 2 negative real roots exist for given f(x) = 0..

So in all we have 6 real roots for the given equation..

Hence D) should be the correct answer

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but Descartes' rule of signs only tells maximum possible (+ve) or (-ve) root, don't tell anything about same or different root.
$\left ( x-2 \right )^{2}$ , which can be written as $x^{2}-2x+4$ , has 2 sign changes. Therefore, the polynomial has 2 positive roots.and here both positive root are same.

putting -x in f(x), f(-x)=$x^{2}+2x+4$ no sign change, so no -ve root.
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Yes we cannot be sure that all 6 root be distinct
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is there any other method to solve.?
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we can use Newton-Raphson Method to find the roots.

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