IIT JEE , AIEEE Forum - Mathematics , Algebra

ramyadiamond

1. If p,q,r,s

R, then (x²+px+3q)(-x²+rx+q)(-x²+sx-2q)=0 has

(a) 6 real roots

(b)at least two real roots

(c) 2 real and 4 imaginary roots

(d) 4 real and 2 imaginary roots

I'm having problems in solving typical problems of this nature. Plz tell me abt any approach to it. Or any way of finding the no. of real roots for any given equation.

2. If alpha is a root of 4x²+2x-1=0 and f(x)=4x³-3x+1, then 2{f(alpha)+1}=

(a)0

(b) -1

(c) 1

(d) none

Plzz help me

rishikesh_anshu

AT least two real root

D₁= p²-12q for    x²+px+3q             --------1
D₂= r²+4q    for -x²+rx+q            --------2
D₃= s²-8q              for    -x²+sx-2q              ----------3
we know for imaginary D<0
case 1: q<0   D₁>0 D₃>0    D₂ may or may not greater than 0 hence atleast 4 real solution
case 2:q>0      D₁ D₃    may or may not greater than 0 D_2>0
hence atleast 2 real solution

taking common answer atleast two real solution

ramyadiamond

vish0001

ecellent answer rishikesh....
for the second one.... what i think is that u will get imaginary roots.. so.. what it seems to be till now is that what u can do is substitute it in the given expression !
aneways.. i am trying !

rishikesh_anshu

f(x)=4x³-3x+1=4x³+2x²-2x²-x+x-3x+1=(4x³+2x²-x)+(-2x²-2x+1)
=x(4x²+2x-1)-(4x²+2x-1)+2x²As

is root of 4x²+2x-1 so
f(

)=

(0)-0+2

²=2

²
2{f(alpha)+1}=2(2

²+1)=(4

²+2

-1)-2

+3
=0-2

+3
=3-2

=(-2+

20)/8 or (-2-

20)/8

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