CAT 2008: Quantitative Questions a Day 1 to 50 - The discussions

[Aarav]

Originally Posted by thebornattitude

sum of roots p+q+r=-1
pq+pr+qr= -3
pqr = -2

1'st and 3'rd has been solved perfectly above, will go with the 2'nd one.. please be with me, as it is a long solution

We will form a new euation with 2-p, 2-q, 2-r as roots of the equation
let them be x,y,z

Sum of the roots ->x+y+z= 6-(p+q+r)= 7............(i)

xy+yz+xz= (2-p)(2-q)+ (2-q)(2-r) + (2-r)(2-p),
solving we get
12- 4(p+q+r) + (pq+pr+qr)= 13...........(ii)

xyz= (2-p)(2-q)(2-r),,solving we get
8- 4(p+q+r) +2(pq+pr+qr) -pqr= 8.................(iii)

now we will simplify (x+y+z) ^3.... calculating it we get
= x^3 +y^3 +z^3 +3{ xy(x+y) + yz(y+z) + xz(x+z) + 2xyz}
-> from (i),,,we can get
(x+y+z)^3 = x^3 +y^3 + z^3 +3( xy(7-z) + yz(7-x) + xz(7-y) + 2xyz)
solving we get
(x+y+z)^3= x^3 + y^3 +z^3 +3( 7(xy+xz+yz)- xyz)
putting the values from (i)(ii)(iii), we get

x^3 +y^3 +z^3 = 94
hence (2-p)^3 + (2-q)^3 + (2-r)^3 = 94 ..answer

Valliant attempt

But, this can be done in 3 lines -> would any one else want to have a go at this?

[deep@k]

Originally Posted by thebornattitude

sum of roots p+q+r=-1
pq+pr+qr= -3
pqr = -2

1'st and 3'rd has been solved perfectly above, will go with the 2'nd one.. please be with me, as it is a long solution

We will form a new euation with 2-p, 2-q, 2-r as roots of the equation
let them be x,y,z

Sum of the roots ->x+y+z= 6-(p+q+r)= 7............(i)

xy+yz+xz= (2-p)(2-q)+ (2-q)(2-r) + (2-r)(2-p),
solving we get
12- 4(p+q+r) + (pq+pr+qr)= 13...........(ii)

xyz= (2-p)(2-q)(2-r),,solving we get
8- 4(p+q+r) +2(pq+pr+qr) -pqr= 8.................(iii)

now we will simplify (x+y+z) ^3.... calculating it we get
= x^3 +y^3 +z^3 +3{ xy(x+y) + yz(y+z) + xz(x+z) + 2xyz}
-> from (i),,,we can get
(x+y+z)^3 = x^3 +y^3 + z^3 +3( xy(7-z) + yz(7-x) + xz(7-y) + 2xyz)
solving we get
(x+y+z)^3= x^3 + y^3 +z^3 +3( 7(xy+xz+yz)- xyz)
putting the values from (i)(ii)(iii), we get

x^3 +y^3 +z^3 = 94
hence (2-p)^3 + (2-q)^3 + (2-r)^3 = 94 ..answer

It is correct... but a bit lenghty ...

ny body having another approach....

[nitin360]

If r is the root of f(x) = x^4 + ax^3 - 6x^2 - ax + 1 = 0, then which among the following is also a root of f(x) = 0?

(1) -1/r (2) (1+r)/(1-r) (3) (r-1)/(r+1) (4) All of these (5) Exactly two of these

my take ...
f(x) = x^4 + ax^3 - 6x^2 - ax + 1 = 0

now putting x=-1/r
and we get the same eqn
so -1/r is the root of eqn.
again prouduct of roots =1
suppose other are a and b
so r*-1/r*a*b=1

so b=-1/a
taking sum of the roots two at time =-6
so on solving we get
a,b=(1+r)/(1-r) and (r-1)/(r+1)
so ans is option 4

[thebornattitude]

Originally Posted by Aarav

Valliant attempt

But, this can be done in 3 lines -> would any one else want to have a go at this?

We can also do it by simply expanding the equation (2-p)^3.......
just did it now....and getting the same answer....

[sudeepdeb]

Originally Posted by deep@k

It is correct... but a bit lenghty ...

ny body having another approach....

About the length, if we substitute x by (2-x) and expand the equation we readily get the sum,products of roots.

Now finding, x^3+y^3+z^3 is same by finding it from (x+y+z)^3, i think this step can't be reduced.

[asd_dah]

IF r is the root of f(x) = x^4 + ax^3 - 6x^2 - ax + 1 = 0, then which among the following is also a root of f(x) = 0?

(1) -1/r (2) (1+r)/(1-r) (3) (r-1)/(r+1) (4) All of these (5) Exactly two of these

here it goes ...
f(x) = x^4 + ax^3 - 6x^2 - ax + 1 = 0

now putting x=-1/r
and we get the same eqn hence -1/r is the root of eqn.
again prouduct of roots from the above eqn.=1
suppose other are a and b
so r*-1/r*a*b=1

again the rest given roots satisfy this condition
a,b=(1+r)/(1-r) and (r-1)/(r+1)
so ans is option 4

[Aarav]

Originally Posted by sudeepdeb

1) About the length, if we substitute x by (2-x) and expand the equation we readily get the sum,products of roots.

2) Now finding, x^3+y^3+z^3 is same by finding it from (x+y+z)^3, i think this step can't be reduced.

The 1st is Ok - but not the 2nd. Replace y = 2-x => x = 2-y and get a cubic in y.
The asked expression would be (a+b+c)((a+b+c)^2 - 3(ab + bc + ca)) + 3abc if a, b, c are the roots of equation in y.

I will put another challenge for you and solve this fully now.

Compute
(2-p^2)^3 + (2-q^2)^3 + (2-r^2)^3

[siddooba]

My try at the second ..

a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - bc - ca)
a^3 + b^3 + c^3 = (a+b+c)((a+b+c)^2 - 3(ab+bc+ca)) + 3 abc

(a+b+c) = (6-(p+q+r)) = 7
(abc)= (2-p)(2-q)(2-r) = (8-4(p+q+r)+2(pq+qr+rq)+pqr)) = 8-4*-1+2(-3)-2 = 6
(ab+bc+ca) = (2-p)(2-q)+.... = 12+(pq+qr+pr)-4(p+q+r)=12-3-4(-1) = 13

Substituting

= 7(49-39) + 3*6
= 70 + 18
= 88

[siddooba]

Originally Posted by Aarav

The 1st is Ok - but not the 2nd. Replace y = 2-x => x = 2-y and get a cubic in y.

I do not understand these lines about replacing

[Aarav]

I think very few have bothered to read the official solution to today's problem: why not use substitution and get our answer than solving long expressions?