ax ˛ + bx + c = 0 has no rational roots if a, b, c are odd integers (True/False)
Let roots be rational a/b, c/d
(x - a/b)(x - c/d) = 0
x^2 - (a/b + c/d)x + (ac)/(bd) = 0
x^2 - ((ad+bc)/(bd))x + (ac)/(bd) = 0
(bd)x^2 - (ad+bc)x + ac = 0
a'=bd, b' =-(ad+bc), c'=ac
odd=odd, odd=even NOT POSSIBLE, odd=odd
so TRUE
@chillfactor said:ax ˛ + bx + c = 0 has no rational roots if a, b, c are odd integers (True/False)
True.
For last question, its true
All three vertices of an equilateral triangle in a Cartesian plane can have integral coordinates. (True/False)
@chillfactor said:For last question, its trueAll three vertices of an equilateral triangle in a Cartesian plane can have integral coordinates. (True/False)
Let one vertex be at origin, Any plane can be shifted so that one point lie on origin wrt new shifted cordinates.
A(0,0) let B(0,x) then C(x/2,y)
AC=AB
y^2+x^2/4=x^2
y=sqrt(3)x/2, So all coordinates can't be integer As here sqrt(3) incurred.
SO FALSE
A(0,0) let B(0,x) then C(x/2,y)
AC=AB
y^2+x^2/4=x^2
y=sqrt(3)x/2, So all coordinates can't be integer As here sqrt(3) incurred.
SO FALSE
True or false:
1) 5^12 + 2^10 is a composite number .
2) (5^125 -1)/(5^25 -1) is a composite number.
@chillfactor said:For last question, its trueAll three vertices of an equilateral triangle in a Cartesian plane can have integral coordinates. (True/False)
False.
lets consider the coordinates of triangle be (0,0),(x,y),(w,z)
all x,y,z,w are integers.
Area of Triangle = 1/2*(0(y-0)+x(z-0)+w(0-y))
=> 1/2*(x*z-w*y) --------------------------------------------> [1]
now area of equi. traingle = (sqrt(3)*a^2)/4 -------------------> [2]
for any integral value of w,x,y,z eqn 1 comes out to be integer. but area of equi triangle from eqn 2 is irrational.
hence FALSE
@chillfactor said:For last question, its trueAll three vertices of an equilateral triangle in a Cartesian plane can have integral coordinates. (True/False)
let one vertex be at origin A(0,0)
then second vertex would lie on line y=(root3)x
and third vertex would lie on line y=-(root3)x
no coordinates combination (x,y) lying on above lines will be integer
so for possibility in question:FALSE
then second vertex would lie on line y=(root3)x
and third vertex would lie on line y=-(root3)x
no coordinates combination (x,y) lying on above lines will be integer
so for possibility in question:FALSE
@chillfactor said:For last question, its trueAll three vertices of an equilateral triangle in a Cartesian plane can have integral coordinates. (True/False)
False. The height of an equilateral triangle is (rt3)/2. therefore . ..
@19rsb said:let one vertex be at origin A(0,0)then second vertex would lie on line y=(root3)xand third vertex would lie on line y=-(root3)xno coordinates combination (x,y) lying on above lines will be integerso for possibility in question:FALSE
bahut badiya bhai bilkul sahi hai 
@kingsleyx said:True or false:1) 5^12 + 2^10 is a composite number .2) (5^125 -1)/(5^25 -1) is a composite number.
5^12 + 2^10 = (5^6 + 2^5)^2 - {(5^6)(2^6)} = (5^6 + 2^5)^2 - {(5^3)(2^3)}^2
= (5^6 + 2^5 - 10^3)(5^6 + 2^5 + 10^3)
So, composite
(5^125 - 1)/(5^25 - 1) = 5^100 + 5^75 + 5^50 + 5^25 + 1
x^4 + x^3 + x^2 + x + 1, where x = 5^25
= (x^2 + 3x + 1)^2 - (5x^3 + 10x^2 + 5x)
= (x^2 + 3x + 1)^2 - 5x(x + 1)^2
For x = 5^25, this expression will become a^2 - b^2 and can be factorized as (a + b)(a - b)
So, composite
For last question (equilateral triangle), statement is false.
@kingsleyx said:True or false:1) 5^12 + 2^10 is a composite number .2) (5^125 -1)/(5^25 -1) is a composite number.
for 1
5^12 + 2^10 = (5^6)^2 + (2^5)^2 = (5^6+2^5)^2 - 2*5^6*2^5
= (5^6+2^5)^2 - 5^6*2^6 = (5^6+2^5)^2 - 10^6 = (5^6+2^5)^2 - (10^3)^2
= (5^6+2^5-10^3)*(5^6+2^5+10^3)
hence composite
@chillfactor said:5^12 + 2^10 = (5^6 + 2^5)^2 - {(5^6)(2^6)} = (5^6 + 2^5)^2 - {(5^3)(2^3)}^2= (5^6 + 2^5 - 10^3)(5^6 + 2^5 + 10^3)So, composite(5^125 - 1)/(5^25 - 1) = 5^100 + 5^75 + 5^50 + 5^25 + 1x^4 + x^3 + x^2 + x + 1, where x = 5^25= (x^2 + 3x + 1)^2 - (5x^3 + 10x^2 + 5x)= (x^2 + 3x + 1)^2 - 5x(x + 1)^2For x = 5^25, this expression will become a^2 - b^2 and can be factorized as (a + b)(a - b)So, compositeFor last question (equilateral triangle), statement is false.
damn!!! i m so slow in typing 😞
@gautam22 said:Prove that for any natural N, 1000^N - 1 cannot be a divisor of 1978^N - 1
Another way to solve this question:-
If A divides B, where B > A, then A will also divide (B - A)
Suppose 1000^N - 1 divides 1978^N - 1, then
1000^N - 1 will also divide 1978^N - 1000^N
1978^N - 1000^N = (2^N)(989^N - 500^N)
=> 1000^N - 1 will divide 989^N - 500^N (as 2^N is even and 1000^N - 1 is odd)
But its not possible as 1000^N - 1 > 989^N - 500^N
So, 1000^N - 1 will not divide 1978^N - 1
@Tina_angel said:2 ques on probability..1. From a group of 5 men and 4 women a committee is to be formed so that in any of these committees there are more women than men. How many different committees can be formed?2. 15 ppl sit round a circular table. What are the odds against two particular ppl sitting together?
5C3*4C4+5C2*4C4+5C1*4C4+5C0*4C4+4C3*5C2+4C3*5C1+4C3*5C0+4C2*5C1+4C2*5C0+4C1*5C0
10+10+5+1+40+20+4+30+6+4
130
2->
7!*2^7/15!
2^7/15*14*13*12*11*10*9*8
@chillfactor said:ax ˛ + bx + c = 0 has no rational roots if a, b, c are odd integers (True/False)
a,b,c=2k+1 or -(2k+1)
x1+x2=-b/a
x1+x2=1 or x1+x2=-1
x1*x2=c/a
x1*x2=1 or x1*x2=-1
Case1-> x^2-x+1=0
Droots are complex
Case2-> x^2-x-1=0
roots are 1+/-rt5/(2) roots are irrational
Case->3 x^2+x+1=0
Droots are complex
Case->4 x^2+x-1=0
Roots are -1+/-rt5/(2) =>roots are irrational
No ,rational roots
@chillfactor said:For last question, its trueAll three vertices of an equilateral triangle in a Cartesian plane can have integral coordinates. (True/False)
I know one combination
(0,a),(0,-a),(2a,0)