difference between 2 positive numbers is 3 and the sum of their squares is 117 Declare variables for each of the two numbers: [LIST] [*]Let the first variable be x [*]Let the second variable be y [/LIST] We're given 2 equations: [LIST=1] [*]x - y = 3 [*]x^2 + y^2 = 117 [/LIST] Rewrite equation (1) in terms of x by adding y to each side: [LIST=1] [*]x = y + 3 [*]x^2 + y^2 = 117 [/LIST] Substitute equation (1) into equation (2) for x: (y + 3)^2 + y^2 = 117 Evaluate and simplify: y^2 + 3y + 3y + 9 + y^2 = 117 Combine like terms: 2y^2 + 6y + 9 = 117 Subtract 117 from each side: 2y^2 + 6y + 9 - 117 = 117 - 117 2y^2 + 6y - 108 = 0 This is a quadratic equation: Solve the quadratic equation 2y2+6y-108 = 0 With the standard form of ax2 + bx + c, we have our a, b, and c values: a = 2, b = 6, c = -108 Solve the quadratic equation 2y^2 + 6y - 108 = 0 The quadratic formula is denoted below: y = -b ± sqrt(b^2 - 4ac)/2a [U]Step 1 - calculate negative b:[/U] -b = -(6) -b = -6 [U]Step 2 - calculate the discriminant ?:[/U] ? = b2 - 4ac: ? = 62 - 4 x 2 x -108 ? = 36 - -864 ? = 900 <--- Discriminant Since ? is greater than zero, we can expect two real and unequal roots. [U]Step 3 - take the square root of the discriminant ?:[/U] ?? = ?(900) ?? = 30 [U]Step 4 - find numerator 1 which is -b + the square root of the Discriminant:[/U] Numerator 1 = -b + ?? Numerator 1 = -6 + 30 Numerator 1 = 24 [U]Step 5 - find numerator 2 which is -b - the square root of the Discriminant:[/U] Numerator 2 = -b - ?? Numerator 2 = -6 - 30 Numerator 2 = -36 [U]Step 6 - calculate your denominator which is 2a:[/U] Denominator = 2 * a Denominator = 2 * 2 Denominator = 4 [U]Step 7 - you have everything you need to solve. Find solutions:[/U] Solution 1 = Numerator 1/Denominator Solution 1 = 24/4 Solution 1 = 6 Solution 2 = Numerator 2/Denominator Solution 2 = -36/4 Solution 2 = -9 [U]As a solution set, our answers would be:[/U] (Solution 1, Solution 2) = (6, -9) Since one of the solutions is not positive and the problem asks for 2 positive number, this problem has no solution

Free Difference of Two Squares Calculator - Factors a difference of squares binomial in the form a² - b² or multiplies 2 binomials through in the form (ax + by)(ax - by).

Free Factoring and Root Finding Calculator - This calculator factors a binomial including all 26 variables (a-z) using the following factoring principles:
* Difference of Squares
* Sum of Cubes
* Difference of Cubes
* Binomial Expansions
* Quadratics
* Factor by Grouping
* Common Term
This calculator also uses the Rational Root Theorem (Rational Zero Theorem) to determine potential roots
* Factors and simplifies Rational Expressions of one fraction
* Determines the number of potential positive and negative roots using Descarte’s Rule of Signs

Find two consecutive positive integers such that the difference of their square is 25. Let the first integer be n. This means the next integer is (n + 1). Square n: n^2 Square the next consecutive integer: (n + 1)^2 = n^2 + 2n + 1 Now, we take the difference of their squares and set it equal to 25: (n^2 + 2n + 1) - n^2 = 25 Cancelling the n^2, we get: 2n + 1 = 25 [URL='https://www.mathcelebrity.com/1unk.php?num=2n%2B1%3D25&pl=Solve']Typing this equation into our search engine[/URL], we get: n = [B]12[/B]

Take an integer n. The next alternate consecutive integer is n + 2 Subtract the difference of the squares: (n + 2)^2 - n^2 n^2 + 4n + 4 - n^2 n^2 terms cancel, we get: 4n + 4 Factor out a 4: 4(n + 1) If n is odd, n + 1 is even. 4 * even is always even If n is even, n + 1 is odd. 4 * odd is always odd Since both cases are even, we've proven our statement. [MEDIA=youtube]J_E9lR5qFY0[/MEDIA]

Take an integer n. The next consecutive integer is n + 1 Subtract the difference of the squares: (n + 1)^2 - n^2 n^2 + 2n + 1 - n^2 n^2 terms cancel, we get: 2n + 1 2 is even. For n, if we use an even: we have even * even = Even Add 1 we have Odd 2 is even. For n, if we use an odd: we have even * odd = Even Add 1 we have Odd Since both cases are odd, we've proven our statement. [MEDIA=youtube]RAi0HbH5bqc[/MEDIA]

The difference between the squares of two consecutive numbers is 141. Find the numbers Take two consecutive numbers: n- 1 and n Given a difference (d) between the squares of two consecutive numbers, the shortcut for this is: 2n - 1 = d Proof of this: n^2- (n - 1)^2 = d n^2 - (n^2 - 2n + 1) = d n^2 - n^2 + 2n - 1 = d 2n - 1 = d Given d = 141, we have 2n - 1 = 141 Add 1 to each side: 2n = 142 Divide each side by 2: 2n/2 = 142/2 n = [B]71[/B] Therefore, n - 1 = [B]70 Our two consecutive numbers are (70, 71)[/B] Check your work 70^2 = 4900 71^2 = 5041 Difference = 5041 - 4900 Difference = 141 [MEDIA=youtube]vZJtZyYWIFQ[/MEDIA]

The first significant digit in any number must be 1, 2, 3, 4, 5, 6, 7, 8, or 9. It was discovered that first digits do not occur with equal frequency. Probabilities of occurrence to the first digit in a number are shown in the accompanying table. The probability distribution is now known as Benford's Law. For example, the following distribution represents the first digits in 231 allegedly fraudulent checks written to a bogus company by an employee attempting to embezzle funds from his employer. Digit, Probability 1, 0.301 2, 0.176 3, 0.125 4, 0.097 5, 0.079 6, 0.067 7, 0.058 8, 0.051 9, 0.046 [B][U]Fradulent Checks[/U][/B] Digit, Frequency 1, 36 2, 32 3, 45 4, 20 5, 24 6, 36 7, 15 8, 16 9, 7 Complete parts (a) and (b). (a) Using the level of significance α = 0.05, test whether the first digits in the allegedly fraudulent checks obey Benford's Law. Do the first digits obey the Benford's Law?
Yes or No Based on the results of part (a), could one think that the employe is guilty of embezzlement? Yes or No Show frequency percentages Digit Fraud Probability Benford Probability 1 0.156 0.301 2 0.139 0.176 3 0.195 0.125 4 0.087 0.097 5 0.104 0.079 6 0.156 0.067 7 0.065 0.058 8 0.069 0.051 9 0.03 0.046 Take the difference between the 2 values, divide it by the Benford's Value. Sum up the squares to get the Test Stat of 2.725281277 Critical Value Excel: =CHIINV(0.95,8) = 2.733 Since test stat is less than critical value, we cannot reject, so [B]YES[/B], it does obey Benford's Law and [B]NO[/B], there is not enough evidence to suggest the employee is guilty of embezzlement.