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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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2 Answers

Best answer
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8 votes

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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1 votes
1 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

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