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What is the product of two consecutive integers that equals 132?
Let the two consecutive integers be x and x+1. The product of these two integers is x(x+1). Setting this equal to 132 gives us the equation x(x+1) = 132. By expanding the left side of the equation, we get x^2 + x = 132. Rearranging terms, we have x^2 + x - 132 = 0. This is a quadratic equation that can be factored as (x+12)(x-11) = 0. Therefore, the two consecutive integers are 11 and 12, and their product is 132.
What is x times -8x?
The product of x and -8x is found by multiplying the coefficients (numbers) and adding the exponents. In this case, x has an understood exponent of 1, so the product is -8x^2. This result comes from multiplying -8 by 1 and adding the exponents of x (1+1=2).
What is 672 as a product of its prime?
2 x 2 x 2 x 2 x 2 x 3 x 7 = 672
The product of 2 consecutive integers is 156?
12 x 13 = 156 How to find the answer: If x is the first integer, then x+1 is the next consecutive integer. so x(x+1) = 156 or x^2 + x -156 =0 solve the quadratic equation for x.
Find two consecutive positive integers whose product is 34 more than their sum?
I don't think there's a solution to this problem. Let's work it through.Two consecutive integers: x & y. So y = x + 1.Product is 34 more than sum. Product (x*y) = sum(x+y) +34Substitute [y=x+1] into second equation: x*(x+1) = x + (x+1) + 34x2 + x = x + x + 1 + 34 --> x2 + x - 2*x - 35 = 0 --> x2 - x - 35 = 0 (quadratic)In the quadratic formula: a = 1, b = -1, c = -35, so we have (1 ± sqrt[1 - 4*(-35)])/(2*1) = (1 ± sqrt(141))/2; sqrt(141) is irrational, so the solution is not integers. Someone else can maybe check my work.
