IIT JEE , AIEEE Forum - Mathematics , Algebra - x^5-17x+a=0 find values of x so that the eqn has real roots

tanushreeroy

x^5-17x+a=0 find values of x so that the eqn has real roots

arpitddn

a=5

b=-17

c=a

b²-4ac>0

(-17)²-4*5*a>0

289>20a

14.45>a

a>14.45

arpitddn

Please tell me if I am correct

srujana

@ above

guess u were trying to use b^2-4ac >=0

ur ans wud have been crt if it was a polynomial of degree 2....

RyuAmakusa

if the q is "find the value of x" then there are many values.else if the q is "find the value of a" even then u have infinitely many roots bcos keep on substuting diff real values of x u get diff values of a for which the eq has real root that is the value which u substitued

srujana

exactly....

here r my assumptions n my approach...

at the frst place it shd be x as pointed out....n the q shd have been 'atleast 1 real root'

n then...assume the graph x^5-17x=0 can be drawn easily...

(refer to the attached pic)

wat u need is tat 'pull' the graph(the red dot) such that it touches the x axis...then it has exactly 2 real roots...n wen u pull it further down...it ll have only 1 root(real)....so tat shd be the 'limiting' value....

srujana

Heres another method which doesnt require the graph...

f(x)=x^5-17x+a

equating the first derivative to 0...(to find out whr the slope is 0...or to find out $f_{max}(x)$ )

$|x|=\left({\frac{17}{5}}\right)^{\frac{1}{4}}$

The corresponding y value will be, $f\left(\left({\frac{17}{5}}\right)^{\frac{1}{4}}\right)$ = 18.4675

For $a \le$ $f\left(\left({\frac{17}{5}}\right)^{\frac{1}{4}}\right)$ ,the equation has exactly 1 root $\approx$ 2

lly for $a \ge$ $f\left(\left({\frac{17}{5}}\right)^{\frac{1}{4}}\right)$ ,the equation has exactly 1 root $\approx$ -2

RyuAmakusa

now i did not get

"The corresponding y value will be, ..."

this step how can we find the value of f(something) if we do not know the value of a.

thedumbheadwithnobrain

pramod6990

dude...but AM > GM is only valid for Positive numbers naa....how do u assure that all the roots are positive yaar....cant u do it this way....we haveF' =0 at 2 pts...+(17/5)^1/4 and ...-(17/5)^1/4

so we realise that we can at max have just 3 distinct roots and the 2 are repeated... so for all roots real...the basic criteria wud probably be F(+(17/5)^1/4 ) . F(-(17/5)^1/4 ) < 0

solving which we may et the reqd. condition on a....

evn i am not sure of my method...but bhai...plz think of this...(AM>GM)

kria

b^2-4ac>0 for real roots

x^5...............?

wll

5x^4-17=0

x^4=17/5

x=+-(17/5)^1/4...........

val of x in dis interval hav real roots..................may be not sure

hsbhatt

By any chance did you mean rational roots like some other problem in this forum. I suspect this because:

1. Since this equation is of odd degree with real coefficients, it certainly has one real root (the non-real roots come in conjugate pairs and hence number of non-real roots is always even in this case)

2. The equation cannot have all roots real as the sum of roots and the sum of product of roots taken two at a time are both zero (coeffs of x⁴ and x³ respectively). This gives sum of squares of roots = 0 which is not possible.

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