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+4 votes

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
asked in Set Theory & Algebra by Boss (17.6k points)
edited by | 290 views
@Tesla!, Hows it done for powers less than 6 or even lesser? All basic algebra of skul days. :).
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.

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!

@set2018 see answer by Heisenberg it is correct explanation


I am asking how 1 real root possible with P'(x)

2 Answers

+5 votes
Best answer

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

answered by Active (2.5k points)
edited by

 answer is A

but can you explain with a better approach?

0 votes

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)


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

answered by Veteran (100k points)
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.
Yes we cannot be sure that all 6 root be distinct
is there any other method to solve.?
we can use Newton-Raphson Method to find the roots.

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