Solve it!!!
How many real solutions for x are there such that
x 3 − 3 x = x + 2 ?
The answer is 3.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
3 solutions
How can a degree 6 (obtained after squaring both sides) polynomial have 3 real solutions ?
If we solve the equation:
x 3 − 3 x = x + 2
We get 3 values of x .
So, the answer is 3
actually it forms a cubic hence it can have 3 possible values of x as a cubic has maximum 3 roots
3 can never be the answer. the equation is actually of degree 6 (if you square both sides). As complex roots for an equation always exist in conjugate pairs there can only be an even number of complex roots which means there has to be an even number of real roots for a degree 6 equation. I got 2 real roots one lying between -root(3)) an 0 and the other as 2.
Log in to reply
By squaring, you introduce extraneous roots. The only roots are 2, -1,618, and 1.24698. Ed Gray
Log in to reply
I made a typo in my previous comment. the roots are x = 2, x + -1.618, and x = -.44504. Ed Gray
That's really wrong. It has 6 roots. The roots are 2 , -1.618 , 0.61803 , -1.8019 , -0.44504 , 1.24698
Log in to reply
If you evaluate x^3 - 3x and sqrt(x + 2) for your solutions, you will find that only x = 2, x = -1.618, and x = 1.24698 result in an equality. Remember that sqrt(x + 2) must be >0, so if x^3 - 3x is negative, they are extraneous roots. Ed Gray
Log in to reply
I made a typo in my comment. The roots should be x = 2, x = -1.618, and x = -.44504. Ed Gray
it is 3 degree polynomial. @Ramasubramaniyan Gunasridharan . degree is determined by highest power of x in polynomial without any modification. eq. by squaring ,history of sqrt.(x+2) is deleted.for eg. sqrt.(4)=2,but sq 2=sq -2=4.hence always be cautious before squaring any term.it might led to an extra solution which was ignored by doing this operation. by the way one can simply find the solution by plotting the two functions.R.H.S. is an odd function with roots 0,1.732,-1.732.while L.H.S. is a parabola with vertex(-2,0).we find only 3 solutions.now you will realize where were you mistaken.
Log in to reply
But can we say an expression as a polynomial if the degree of x is not a whole number. This is what we study in secondary school.
if we see equation - x^6-〖6x〗^4+〖9x〗^2-x^1-2=0 it has 2 which gives us the cofactors values(+1,-1,+2,-2) but only (+2) satisfy equation therefore answer is 1.
Hey How could it be a level 3
x = c o s θ and we will get 3 difference answers.
we can take x=cosa only when domain of the is in [-1,1]
Consider this. y = x 3 − 3 x . . . ( R e d )
y = x + 2 . . . ( B l u e )
imgur
We can see that the graphs intersect at 3 points. Therefore, there are 3 real solutions for x .