Question Bank  Algebra  Hemant Malhotra

@hemant_malhotra can we use same exponent multiple times ? eg. you have a = 1 in multiple cases.

@hemant_malhotra the final expression will have power of 3m where m is whole number till 200. So total number of terms is 201 ?

[Solution by @GauravKapoor ]
We have a^2 + ab + b^2 = 1
It can be written as
a(a + b) + b^2 = 1       (i)
and b(a + b) + a^2 = 1     (ii)
Adding the two equations we get:
(a + b)^2 + a^2 + b^2 = 2
Now minimum value of (a+b)^2 = 0
So different values of (a,b) will be
(1, 0),
(0, 1),
(1, –1),
(–1, 1),
(–1, 0),
(0, –1)So 6 pairs will be there.

@VikrantGarg  Yes, no real roots possible.

Answer is 121.
Method 1 by rishabh tyagi
Boundary points are
(8,9),(9,8),(10,7)..(16,1)==>9 points
(2,1),(1,2)...(6,9)==>9 points
(2,0),(1,0)...(16,0)==>19 points
(3,0),(7,10),(17,0)==> 3 points
Total 40Method 2 by @hemant_malhotra :
x  7 + y + k =10
12c2= 66
as x7 can take ve values 66 * 2 =132
now subtract cases when x  7 = 0 so y < = 10 so 11 values will be removed so 13211=121Method 3 by @hemant_malhotra
x  7 + y < = 10
let x  7 > = 0
x7 + y < = 10
so x + y < = 17
here x > = 7
so put x = 7 + a
so 7 + a + y < = 17
so a + y < = 10
so when a = 0 then y < = 10 so 11 values
when a=1 then y < = 9 so 10 values
.
.
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when a = 7 then 4 values
when a = 8 then y < = 2
when a = 9 then y < = 1
so 2 values
when a = 10 then y < = 0 so 1 values1 + 2 + 3 ... + 11 = 11 * 6 = 66 values
now when 0 < = x < = 6
so x + 7 + y < = 10
so x + y < = 3
y  x < = 3
so y < = x + 3
so when x = 0 then y < = 3 so 4 values
when x = 1 then 5 values
tlll x = 6 where 10 values
so 4 + 5 + 6 + 7 + 8 + 9 + 10 = 49now when x < 0
so x + 7 + y < = 10
so x + y < = 3
so y < = x + 3
when x = 1 then y < = 2 so 3 values
when x = 2 then y < = 1 so 2 values
x = 3 y < = 0 so 1 values so 6 value totall66 + 49 + 6 = 121 values

@hemant_malhotra said in Question Bank  Algebra  Hemant Malhotra:
If x & y are integers then how many values of ( x , y ) shall satisfy 16 <  x  +  y  < 30
what is the answer sir ?

@hemant_malhotra a,b,c=1,2,3
2×2×2=8

@zabeer sir but here Different Positive intergers is asked

@hemant_malhotra 15/2 by proportionality theorem

@hemant_malhotra 4k+2
so 0

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@hemant_malhotra x=4, y=2, z=4
96

@hemant_malhotra said in Question Bank  Algebra  Hemant Malhotra:
If a^2 = 5a  3 and b^2 = 5b  3 then find the quadratic eqn whose roots are a/b and b/a
3x^219x+3=0

@hemant_malhotra d.......

@hemant_malhotra said in Question Bank  Algebra  Hemant Malhotra:
Let a < b < c be the three real roots of the equation √2014 x^3  4029 x^2 + 2 = 0. Find b(a + c).
@zabeer sir help

@hemant_malhotra (8/3)^4 ( 5) ^5

@hemant_malhotra ...........d............

@hemant_malhotra 20 .....................

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@hemant_malhotra said in Question Bank  Algebra  Hemant Malhotra:
Let a < b < c be the three real roots of the equation √2014 x^3  4029 x^2 + 2 = 0. Find b(a + c).
https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_9