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# Polynomials - Possible pairs of Solutions

Given a quadratic equation, we need to find the possible pairs of integer solutions to the equation under certain constraints.
Published on: Mar 4, 2016
Published in: Education

#### Transcripts - Polynomials - Possible pairs of Solutions

• 1. Polynomials Q18
• 2. Qn: Polynomials (a) 6 (b) 12 (c) 24 (d) 48 How many pairs of integer (a, b) are possible such that a2 – b2 = 288?
• 3. Soln: Polynomials 288 = 25 × 32. So it has 6 × 3 = 18 factors. Or, there are 9 ways of writing this number as a product of two positive integers. Let us list these down. 1 × 288, 2 × 144, 3 × 96, 4 × 72, 6 × 48, 8 × 36, 9 × 32, 12 × 24 and 16 × 18 Now, this is where the question gets interesting. If a, b are integers either a + b and a – b have to be both odd or a+ b and a –b have to be both even. So, within this set of possibilities 1 × 288, 3 × 96 and 9 × 32 will not result in integer values of a, b. So, there are 6 sets of numbers that work for us. How many pairs of integer (a, b) are possible such that a2 – b2 = 288?
• 4. Soln: Polynomials Moving on to these six sets; let us start with one example and see how many possibilities we can generate from this. Let us consider the set 2 × 144. Let us solve for this for a, b being natural numbers first, then we will extend this to integers. When a, b are natural numbers a + b > a – b So, a + b = 144, a – b = 2; a = 73 and b = 71 Now, if a = 73, b = 71 holds good. We can see that a = 73, b = –71 also holds good. a = –73, b = 71 works and so does a = –73, b = –71. How many pairs of integer (a, b) are possible such that a2 – b2 = 288?
• 5. Soln: Polynomials There are 4 possibilities. a = 73, b = 71 a = 73, b = –71 a = –73, b = 71 a = –73, b = –71 So, for each of the 6 products remaining, we will have 4 possibilities each. Total number of (a, b) that will satisfy this equation = 6 × 4 = 24. Answer choice (c) How many pairs of integer (a, b) are possible such that a2 – b2 = 288?