1do the sign change method works in determining the no. of +ve and -ve roots
itz given in many books???
Please solve these problems
1. (x-a)(x-a)(x-a) + (x-b)(x-b)(x-b) + (x-c)(x-c)(x-c) = 0 has
A)All the roots real
B)one real and two imaginary roots
C)three real roots namely x=a,x=b,x=c
D)None
2. a+b+c = 0, then the quadratic equation 3ax.x +2bx +c =0 has roots of range????????
3. If α is a real root of the equation 2xxx - 3xx + 6x + 6 = 0, then value of [α] ,where [] is the greatest integer function
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7 Answers
62Take the 1st derivative...
3{(x-a)2+(x-b)2+(x-c)2} this has no real roots if a≠b≠c
Thus, the orignail function will have only 1 real root unless a=b=c.. in which case they will be 3 repeated roots!
622. a+b+c = 0, then the quadratic equation 3ax.x +2bx +c =0 has roots of range????????
take the function
f(x) = ax3+bx2+cx
f(1)=0
and f(0) = 0
so there is a root in the interval (0,1) (by rollle's theorem)
623. If α is a real root of the equation 2x3 - 3x2 + 6x + 6 = 0, then value of [α] ,where [] is the greatest integer function
derivative is
6x2-6x+6
which is never zero.. so the original function 2x3 - 3x2 + 6x + 6 has only one root.
f(-1)=-2-3-6+6<0
f(0)= 6 > 0
so there is a root between (-1) and 0
so [α] = -1
162yes that also works but i could not figure out how in this problem...
4Thanks for the solutions
but in first soln did u use the converse of the condition that if f(x) has n real roots then f ' (x) has n-1 real roots
Is the converse true!
62The converse is not true.. I have not used the converse either...
A=> B is same as Bc=>Ac
A implies B is equivalent to not B implies not A
that is what i have used