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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
asked in Set Theory & Algebra by Boss (17.6k points)
edited by | 290 views
0
@Tesla!, Hows it done for powers less than 6 or even lesser? All basic algebra of skul days. :).
0
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.
0

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!

0
@set2018 see answer by Heisenberg it is correct explanation
0

@Tesla!  

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
0

 answer is A  https://www.wolframalpha.com/input/?i=x6%E2%88%925x4%2B16x2%E2%88%9272x%2B9%3D0

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)

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

answered by Veteran (100k points)
+3
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.
+1
Yes we cannot be sure that all 6 root be distinct
0
is there any other method to solve.?
0
we can use Newton-Raphson Method to find the roots.

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