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Question Bank - Algebra - Hemant Malhotra

Amar Rajput last edited by @hemant_malhotra

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badalravi last edited by @hemant_malhotra

@hemant_malhotra x=4, y=2, z=4
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Amar Rajput last edited by @hemant_malhotra

@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^2-19x+3=0
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Amar Rajput last edited by @hemant_malhotra

@hemant_malhotra d.......
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Amar Rajput last edited by @hemant_malhotra

@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
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Amar Rajput last edited by @hemant_malhotra

@hemant_malhotra (8/3)^4 ( 5) ^5
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Amar Rajput last edited by @hemant_malhotra

@hemant_malhotra ...........d............
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Amar Rajput last edited by @hemant_malhotra

@hemant_malhotra 20 .....................
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Anurag MN last edited by @hemant_malhotra

@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
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Let the two digit number be represented as ab. a^2 – b^2 = (a – b)(a + b) Now, |a^2 – b^2| will be a prime number only when |a – b| = 1 and (a + b) comes out to be a prime number Two digit numbers in which |a – b| = 1 are 10, 12, 21, 23, 32, 34, 43, 45, 54, 56, 65, 67, 76, 78, 87,89, 98 From these numbers, the numbers in the form of 'ab' in which (a + b) comes out to be a prime number are 12, 21, 23, 32, 34, 43, 56, 65, 67, 76, 89 and 98. There are 12 two digit numbers such that absolute difference of the square of the two digits results in a prime number.
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@sumit-agarwal In such scenarios like finding distinct values of [x^2/n] where x can be from 1, 2, 3 ... n[1^2/n], [2^2/n] ... [(n/2)^2/n] will yield all numbers from 0 to [n/4] (means [n/4] + 1 distinct integers)Then the next set (from [(n/2 + 1)^2/n] till [n^2/n] will be all different integers (means [n/2] distinct integers)So the number of distinct integers would be [n/2] + [n/4] + 1 if n = 100,number of distinct integers would be [100/2] + [100/4] + 1 = 76 if n = 2014,number of distinct integers would be [2014/2] + [2014/4] + 1 = 1511 if n = 13number of distinct integers would be [13/2] + [13/4] + 1 = 10 Just trying to generalize a solution shared by Kamal sir (Quant Boosters - Set 1 - Q2). You can try out with various numbers (may be smaller numbers) so that this can be verified.
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