IIT JEE , AIEEE Forum - Mathematics , Algebra

agent007

Q.1) if a,b and c ae the roots of the equation x³+ 27 = 0, then the quadratic equation whose roots are (c/a)²and (b/a)², is given by ?

Q.2) if a,b be the roots of 4x²-16x+k = 0, where k is real, such that 1<a<2 and 2<b<3 then the number of integral values of k is ?

please don't forget to explain the working also.....

rajat

hey,for the 1st one , i got

quadratic equation is : x^2 - x + 1=0

and for the 2nd one 12 < k < 16

rajat

hey,for the 1st one , i got

quadratic equation is : x^2 - x + 1=0

and for the 2nd one 12 < k < 16

tell me if its ryr then i'll post the solution

agent007

hey rajat,

the first one is not correct, but give your method because you are quite close to the answer...

for the second one, the asks for the number of values, not the interval...

give me the method and i don't might rating your post.... :)

magiclko

x^3 + 27= 0

=> x^3 = -27

=> x = 3 cuberoot (-1)

now -1 = e ^{[ i(2n+1) pi]} whr i= iota

thrfore cuberoot(-1) = e ^{[ i(2n+1) pi] / 3}

frm here a= 3 e^{i pi / 3}

b= 3 e^i pi= -3

c= 3 e^{i 5pi / 3}

so u get the values of a, b and c

rajat

ok , i'll try the 1st one again. maybe i made a silly mistake and for the 2nd one the answer should be 3

magiclko

for the 2nd question

1st condition for k:- D> 0 (as the roots are real)

2nd condition :-

see frm the graph.... f(2) <0, f(1) >0 and f(3) >0

solving all thses u'll get the value of k

ritu_007

hi! the 1st ans.is x²-x-1=0.

rajat

ok

so here are the answers

1 . x^2 + x + 1 =0 roots are w,w^2

2. k can take values 13,14,15

do rate me !!

shash_89

the roots for the 1st que are 3,3w,3w^2.

after you substitute the values to the condn
u get w and w^2.

w+w^2=-1
and w*w^2=1

so the eq is
x^2+x+1.

for the 2nd one 12 < k < 16
do rate me.

agent007

hey magicklo, the method you have applied for the second one is the same as the one i used, but are you sure it is correct??

the answer that k has only 3 values, but how can you know that it doesn't have any more???

magiclko

ofcourse its correct... no doubt about it

all the values of k, will be in this interval only.... id nt knw how many are they!!! i havnt solved the whole problem!!

magiclko

@ shash_89

m afraid u r bit wrong... watever solution u hav done is for x^3 - 27 =0... but here its x^3+ 27 =0...

agent007

oh god, i forgot to rate MAGICLKO!!!!, what a pardonless crime i have done. i will probably have to put my self in exile for the next 14 years!!!!!!

look you guys, ( magiclko and rajat ) we love you even if we forget to rate you OK??? don't take it personally, please!!!

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