quadratic - Polynomials with a maximum term degree as the second degree2 consecutive odd integers such that their product is 15 more than 3 times their sum
2 consecutive odd integers such that their product is 15 more than 3 times their sum. Let the first integer be n. The next odd, consecutive integer is n + 2. We are given the product is 15 more than 3 times their sum: n(n + 2) = 3(n + n + 2) + 15 Simplify each side: n^2 + 2n = 6n + 6 + 15 n^2 + 2n = 6n + 21 Subtract 6n from each side: n^2 - 4n - 21 = 0 [URL='https://www.mathcelebrity.com/quadratic.php?num=n%5E2-4n-21%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']Type this problem into our search engine[/URL], and we get: n = (-3, 7) If we use -3, then the next consecutive odd integer is -3 + 2 = -1. So we have [B](-3, -1)[/B] If we use 7, then the next consecutive odd integer is 7 + 2 = 9. So we have [B](7, 9)[/B]
2 numbers that add up makes 5 but multiplied makes -36 Let the first number be x and the second number be y. We're given two equations: [LIST=1] [*]x + y = 5 [*]xy = -36 [/LIST] Rearrange equation (1) by subtracting y from each side: [LIST=1] [*]x = 5 - y [*]xy = -36 [/LIST] Substitute equation (1) for x into equation (2): (5 - y)y = -36 5y - y^2 = -36 Add 36 to each side: -y^2 + 5y + 36 = 0 We have a quadratic equation. To solve this, we [URL='https://www.mathcelebrity.com/quadratic.php?num=-y%5E2%2B5y%2B36%3D0&pl=Solve+Quadratic+Equation&hintnum=0']type it in our search engine and solve[/URL] to get: y = ([B]-4, 9[/B]) We check our work for each equation: [LIST=1] [*]-4 + 9 = -5 [*]-4(9) = -36 [/LIST] They both check out
2x^2+4x < 3x+6 Subtract 3x from both sides: 2x^2 + x < 6 Subtract 6 from both sides 2x^2 + x - 6 < 0 Using our [URL='http://www.mathcelebrity.com/quadratic.php?num=2x%5E2%2Bx-6&pl=Solve+Quadratic+Equation&hintnum=+0']quadratic calculator[/URL], we get: x < 1.5 and x < -2 When we take the intersection of these, it's [B]x < 1.5[/B]
Free 3 Point Equation Calculator - Forms a quadratic from 3 points that are entered.
A 3 hour river cruise goes 15 km upstream and then back again. The river has a current of 2 km an hour. What is the boat's speed and how long was the upstream journey? [U]Set up the relationship of still water speed and downstream speed[/U] Speed down stream = Speed in still water + speed of the current Speed down stream = x+2 Therefore: Speed upstream =x - 2 Since distance = rate * time, we rearrange to get time = Distance/rate: 15/(x+ 2) + 15 /(x- 2) = 3 Multiply each side by 1/3 and we get: 5/(x + 2) + 5/(x - 2) = 1 Using a common denominator of (x + 2)(x - 2), we get: 5(x - 2)/(x + 2)(x - 2) + 5(x + 2)/(x + 2)(x - 2) (5x - 10 + 5x + 10)/5(x - 2)/(x + 2)(x - 2) 10x = (x+2)(x-2) We multiply through on the right side to get: 10x = x^2 - 4 Subtract 10x from each side: x^2 - 10x - 4 = 0 This is a quadratic equation. To solve it, [URL='https://www.mathcelebrity.com/quadratic.php?num=x%5E2-10x-4%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']we type it in our search engine[/URL] and we get: Speed of the boat in still water =X=5 +- sq. Root of 29 kmph We only want the positive solution: x = 5 + sqrt(29) x = 10.38 [U]Calculate time for upstream journey:[/U] Time for upstream journey = 15/(10.38 - 2) Time for upstream journey = 15/(8.38) Time for upstream journey = [B]1.79[/B] [U]Calculate time for downstream journey:[/U] Time for downstream journey = 15/(10.38 + 2) Time for downstream journey = 15/(12.38) Time for downstream journey = [B]1.21[/B]
Let x be the number. We have: x^2 + x = 30 Subtract 30 from each side: x^2 + x - 30 = 0 Using our [URL='http://www.mathcelebrity.com/quadratic.php?num=x%5E2%2Bx-30%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']quadratic calculator[/URL], we get potential solutions of: [B]x = 5 or x = -6[/B] Check 5: 5 + 5^2 = 5 + 25 = [B]30[/B] Check -6 -6 + -6^2 = -6 + 36 = [B]30[/B]
A flower bed is to be 3 m longer than it is wide. The flower bed will an area of 108 m2 . What will its dimensions be? A flower bed has a rectangle shape, so the area is: A = lw We are given l = w + 3 Plugging in our numbers given to us, we have: 108 = w(w + 3) w^2 + 3w = 108 Subtract 108 from each side: w^2 + 3w - 108 = 0 [URL='https://www.mathcelebrity.com/quadratic.php?num=w%5E2%2B3w-108%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']Type this problem into our search engine[/URL], and we get: w = (9, -12) Since length cannot be negative, w = 9. And l = 9 + 3 --> l = 12 So we have [B](l, w) = (12, 9)[/B] Checking our work, we have: A = (12)9 A = 108 <-- Match!
a rectangle has an area of 238 cm 2 and a perimeter of 62 cm. What are its dimensions? We know the rectangle has the following formulas: Area = lw Perimeter = 2l + 2w Given an area of 238 and a perimeter of 62, we have: [LIST=1] [*]lw = 238 [*]2(l + w) = 62 [/LIST] Divide each side of (1) by w: l = 238/w Substitute this into (2): 2(238/w + w) = 62 Divide each side by 2: 238/w + w = 31 Multiply each side by w: 238w/w + w^2 = 31w Simplify: 238 + w^2 = 31w Subtract 31w from each side: w^2 - 31w + 238 = 0 We have a quadratic. So we run this through our [URL='https://www.mathcelebrity.com/quadratic.php?num=w%5E2-31w%2B238%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']quadratic equation calculator[/URL] and we get: w = (14, 17) We take the lower amount as our width and the higher amount as our length: [B]w = 14 l = 17 [/B] Check our work for Area: 14(17) = 238 <-- Check Check our work for Perimeter: 2(17 + 14) ? 62 2(31) ? 62 62 = 62 <-- Check
A rectangular garden is 5 ft longer than it is wide. Its area is 546 ft2. What are its dimensions? [LIST=1] [*]Area of a rectangle is lw. lw = 546ft^2 [*]We know that l = w + 5. [/LIST] Substitute (2) into (1) (w + 5)w = 546 w^2 + 5w = 546 Subtract 546 from each side w^2 + 5w - 546 = 0 Using the positive root in our [URL='http://www.mathcelebrity.com/quadratic.php?num=w%5E2%2B5w-546%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']quadratic calculator[/URL], we get [B]w = 21[/B]. This means l = 21 + 5. [B]l = 26[/B]
a rocket is propelled into the air. its path can be modeled by the relation h = -5t^2 + 50t + 55, where t is the time in seconds, and h is height in metres. when does the rocket hit the ground We set h = 0: -5t^2 + 50t + 55 = 0 Typing this quadratic equation into our search engine to solve for t, we get: t = {-1, 11} Time can't be negative, so we have: t = [B]11[/B]
An orchard has 378 orange trees. The number of rows exceeds the number of trees per row by 3. How many trees are there in each row? We have r rows and t trees per row. We're give two equations: [LIST=1] [*]rt = 378 [*]r = t + 3 [/LIST] Substitute equation (2) into equation (1) for r: (t + 3)t = 378 Multiply through: t^2 + 3t = 378 We have a quadratic equation. To solve this equation, we [URL='https://www.mathcelebrity.com/quadratic.php?num=t%5E2%2B3t%3D378&pl=Solve+Quadratic+Equation&hintnum=+0']type it in our search engine [/URL]and we get: t = 18 and t = -21 Since t cannot be negative, we get trees per row (t): [B]t = 18[/B]
An orchard has 816 apple trees. The number of rows exceeds the number of trees per row by 10. How many trees are there in each row? Let the rows be r and the trees per row be t. We're given two equations: [LIST=1] [*]rt = 816 [*]r = t + 10 [/LIST] Substitute equation (2) into equation (1) for r: (t + 10)t = 816 t^2 + 10t = 816 Subtract 816 from each side of the equation: t^2 + 10t - 816 = 816 - 816 t^2 + 10t - 816 = 0 We have a quadratic equation. To solve this, we [URL='https://www.mathcelebrity.com/quadratic.php?num=t%5E2%2B10t-816%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']type it in our search engine [/URL]and we get: t = (24, -34) Since the number of trees per row can't be negative, we choose [B]24[/B] as our answer
Free Cubic Equation Calculator - Solves for cubic equations in the form ax3 + bx2 + cx + d = 0 using the following methods:
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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
Divide 73 into two parts whose product is 40 Our first part is x Our second part is 73 - x The product of the two parts is: x(73 - x) = 40 Multiplying through, we get: -x^2 + 73x = 402 Subtract 40 from each side, we get: -x^2 + 73x - 402 = 0 This is a quadratic equation. To solve this, we type it in our search engine, choose "solve Quadratic", and we get: [LIST=1] [*]x = [B]6[/B] [*]x = [B]67[/B] [/LIST]
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Find two consecutive integers if the sum of their squares is 1513 Let the first integer be n. The next consecutive integer is (n + 1). The sum of their squares is: n^2 + (n + 1)^2 = 1513 n^2 + n^2 + 2n + 1 = 1513 2n^2 + 2n + 1 = 1513 Subtract 1513 from each side: 2n^2 + 2n - 1512 = 0 We have a quadratic equation. We [URL='https://www.mathcelebrity.com/quadratic.php?num=2n%5E2%2B2n-1512%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']type this into our search engine[/URL] and get: n = (-27, 28) Let's take the positive solution. The second integer is: n + 1 28 + 1 = 29
Find two consecutive odd integers such that the sum of their squares is 290. Let the first odd integer be n. The next odd integer is n + 2 Square them both: n^2 (n + 2)^2 = n^2 + 4n + 4 from our [URL='https://www.mathcelebrity.com/expand.php?term1=%28n%2B2%29%5E2&pl=Expand']expansion calculator[/URL] The sum of the squares equals 290 n^2 + n^2 + 4n + 4 = 290 Group like terms: 2n^2 + 4n + 4 = 290 [URL='https://www.mathcelebrity.com/quadratic.php?num=2n%5E2%2B4n%2B4%3D290&pl=Solve+Quadratic+Equation&hintnum=+0']Enter this quadratic into our search engine[/URL], and we get: n = 11, n = -13 Which means the two consecutive odd integer are: 11 and 11 + 2 = 13. [B](11, 13)[/B] -13 and -13 + 2 = -11 [B](-13, -11)[/B]
Find two consecutive positive integers such that the sum of their squares is 25. Let the first integer be x. The next consecutive positive integer is x + 1. The sum of their squares equals 25. We write this as:: x^2 + (x + 1)^2 Expanding, we get: x^2 + x^2 + 2x + 1 = 25 Group like terms: 2x^2 + 2x + 1 = 25 Subtract 25 from each side: 2x^2 + 2x - 24 = 0 Simplify by dividing each side by 2: x^2 + x - 12 = 0 Using our [URL='http://www.mathcelebrity.com/quadratic.php?num=x%5E2%2Bx-12%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']quadratic calculator[/URL], we get x = 3 or x = -4. The problem asks for positive integers, so we discard -4, and use 3. This means, our next positive integer is 3 + 1 = 4. So we have [B](3, 4) [/B]as our answers. Let's check our work: 3^2 + 4^2 = 9 + 16 = 25
If 7 times the square of an integer is added to 5 times the integer, the result is 2. Find the integer. [LIST] [*]Let the integer be "x". [*]Square the integer: x^2 [*]7 times the square: 7x^2 [*]5 times the integer: 5x [*]Add them together: 7x^2 + 5x [*][I]The result is[/I] means an equation, so we set 7x^2 + 5x equal to 2 [/LIST] 7x^2 + 5x = 2 [U]This is a quadratic equation. To get it into standard form, we subtract 2 from each side:[/U] 7x^2 + 5x - 2 = 2 - 2 7x^2 + 5x - 2 = 0 [URL='https://www.mathcelebrity.com/quadratic.php?num=7x%5E2%2B5x-2%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']Type this problem into our search engine[/URL], and we get two solutions: [LIST=1] [*]x = 2/7 [*]x= -1 [/LIST] The problem asks for an integer, so our answer is x[B] = -1[/B]. [U]Let's check our work by plugging x = -1 into the quadratic:[/U] 7x^2 + 5x - 2 = 0 7(-1)^2 + 5(-1) - 2 ? 0 7(1) - 5 - 2 ? 0 0 = 0 So we verified our answer, [B]x = -1[/B].
if a number is added to its square, it equals 20. Let the number be an arbitrary variable, let's call it n. The square of the number means we raise n to the power of 2: n^2 We add n^2 to n: n^2 + n It equals 20 so we set n^2 + n equal to 20 n^2 + n = 20 This is a quadratic equation. So [URL='https://www.mathcelebrity.com/quadratic.php?num=n%5E2%2Bn%3D20&pl=Solve+Quadratic+Equation&hintnum=+0']we type this equation into our search engine[/URL] to solve for n and we get two solutions: [B]n = (-5, 4)[/B]
if a number is added to its square, the result is 72. find the number. Let the number be n. We're given: n + n^2 = 72 Subtract 72 from each side, we get: n^2 + n - 72 = 0 This is a quadratic equation. [URL='https://www.mathcelebrity.com/quadratic.php?num=n%5E2%2Bn-72%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']We type this equation into our search engine[/URL], and we get: [B]n = 8 and n = -9[/B]
If the speed of an aeroplane is reduced by 40km/hr, it takes 20 minutes more to cover 1200m. Find the time taken by aeroplane to cover 1200m initially. We know from the distance formula (d) using rate (r) and time (t) that: d = rt Regular speed: 1200 = rt Divide each side by t, we get: r = 1200/t Reduced speed. 20 minutes = 60/20 = 1/3 of an hour. So we multiply 1,200 by 3 3600 = (r - 40)(t + 1/3) If we multiply 3 by (t + 1/3), we get: 3t + 1 So we have: 3600 = (r - 40)(3t + 1) Substitute r = 1200/t into the reduced speed equation: 3600 = (1200/t - 40)(3t + 1) Multiply through and we get: 3600 = 3600 - 120t + 1200/t - 40 Subtract 3,600 from each side 3600 - 3600 = 3600 - 3600 - 120t + 1200/t - 40 The 3600's cancel, so we get: - 120t + 1200/t - 40 = 0 Multiply each side by t: -120t^2 - 40t + 1200 = 0 We've got a quadratic equation. To solve for t, [URL='https://www.mathcelebrity.com/quadratic.php?num=-120t%5E2-40t%2B1200%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']we type this in our search engine[/URL] and we get: t = -10/3 or t = 3. Since time [I]cannot[/I] be negative, our final answer is: [B]t = 3[/B]
If x2 is added to x, the sum is 42. x^2 + x = 42 Subtract 42 from both sides: x^2 + x - 42 = 0 We have a quadratic equation. Using our [URL='http://www.mathcelebrity.com/quadratic.php?num=x%5E2%2Bx-42%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']quadratic equation solver[/URL], we get: [B]x = 6 and x = -7 [/B] Since the problem does not state positive number solutions, they are both answers.
One number is equal to the square of another. Find the numbers if both are positive and their sum is 650 Let the number be n. Then the square is n^2. We're given: n^2 + n = 650 Subtract 650 from each side: n^2 + n - 650 = 0 We have a quadratic equation. [URL='https://www.mathcelebrity.com/quadratic.php?num=n%5E2%2Bn-650%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']We type this into our search engine[/URL] and we get: n = 25 and n = -26 Since the equation asks for a positive solution, we use [B]n = 25[/B] as our first solution. the second solution is 25^2 = [B]625[/B]
Pleasantburg has a population growth model of P(t)=at^2+bt+P0 where P0 is the initial population. Suppose that the future population of Pleasantburg t years after January 1, 2012, is described by the quadratic model P(t)=0.7t^2+6t+15,000. In what month and year will the population reach 19,200? Set P(t) = 19,200 0.7t^2+6t+15,000 = 19,200 Subtract 19,200 from each side: 0.7t^2+6t+4200 = 0 The Quadratic has irrational roots. So I set up a table below to run through the values. At t = 74, we pass 19,200. Which means we add 74 years to 2012: 2012 + 74 = [B]2086[/B] t 0.7t^2 6t Add 15000 Total 1 0.7 6 15000 15006.7 2 2.8 12 15000 15014.8 3 6.3 18 15000 15024.3 4 11.2 24 15000 15035.2 5 17.5 30 15000 15047.5 6 25.2 36 15000 15061.2 7 34.3 42 15000 15076.3 8 44.8 48 15000 15092.8 9 56.7 54 15000 15110.7 10 70 60 15000 15130 11 84.7 66 15000 15150.7 12 100.8 72 15000 15172.8 13 118.3 78 15000 15196.3 14 137.2 84 15000 15221.2 15 157.5 90 15000 15247.5 16 179.2 96 15000 15275.2 17 202.3 102 15000 15304.3 18 226.8 108 15000 15334.8 19 252.7 114 15000 15366.7 20 280 120 15000 15400 21 308.7 126 15000 15434.7 22 338.8 132 15000 15470.8 23 370.3 138 15000 15508.3 24 403.2 144 15000 15547.2 25 437.5 150 15000 15587.5 26 473.2 156 15000 15629.2 27 510.3 162 15000 15672.3 28 548.8 168 15000 15716.8 29 588.7 174 15000 15762.7 30 630 180 15000 15810 31 672.7 186 15000 15858.7 32 716.8 192 15000 15908.8 33 762.3 198 15000 15960.3 34 809.2 204 15000 16013.2 35 857.5 210 15000 16067.5 36 907.2 216 15000 16123.2 37 958.3 222 15000 16180.3 38 1010.8 228 15000 16238.8 39 1064.7 234 15000 16298.7 40 1120 240 15000 16360 41 1176.7 246 15000 16422.7 42 1234.8 252 15000 16486.8 43 1294.3 258 15000 16552.3 44 1355.2 264 15000 16619.2 45 1417.5 270 15000 16687.5 46 1481.2 276 15000 16757.2 47 1546.3 282 15000 16828.3 48 1612.8 288 15000 16900.8 49 1680.7 294 15000 16974.7 50 1750 300 15000 17050 51 1820.7 306 15000 17126.7 52 1892.8 312 15000 17204.8 53 1966.3 318 15000 17284.3 54 2041.2 324 15000 17365.2 55 2117.5 330 15000 17447.5 56 2195.2 336 15000 17531.2 57 2274.3 342 15000 17616.3 58 2354.8 348 15000 17702.8 59 2436.7 354 15000 17790.7 60 2520 360 15000 17880 61 2604.7 366 15000 17970.7 62 2690.8 372 15000 18062.8 63 2778.3 378 15000 18156.3 64 2867.2 384 15000 18251.2 65 2957.5 390 15000 18347.5 66 3049.2 396 15000 18445.2 67 3142.3 402 15000 18544.3 68 3236.8 408 15000 18644.8 69 3332.7 414 15000 18746.7 70 3430 420 15000 18850 71 3528.7 426 15000 18954.7 72 3628.8 432 15000 19060.8 73 3730.3 438 15000 19168.3 74 3833.2 444 15000 19277.2
Quadratic equation hacks using the discriminant Solve x^2- 4x+ 5 using a discriminant: Discriminant is: Discriminant = b^2- 4ac Discriminant = (-4)^2 - 4(1)(5) Discriminant = 16 - 20 Discriminant = -4 When Discriminant < 0, the quadratic has [I][U]no solution [MEDIA=youtube]RogZ3430_8E[/MEDIA][/U][/I]
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Free Quartic Equations Calculator - Solves quartic equations in the form ax4 + bx3 + cx2 + dx + e using the following methods:
1) Solve the long way for all roots and the discriminant Δ
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Sam and Jeremy have ages that are consecutive odd integers. The product of their ages is 783. Which equation could be used to find Jeremy's age, j, if he is the younger man. Let Sam's age be s. Let' Jeremy's age be j. We're given: [LIST=1] [*]s = j + 2 <-- consecutive odd integers [*]sj = 783 [/LIST] Substitute (1) into (2): (j + 2)j = 783 j^2 + 2j = 783 Subtract 783 from each side: j^2 + 2j - 783 = 0 <-- This is the equation to find Jeremy's age. To solve this, [URL='https://www.mathcelebrity.com/quadratic.php?num=j%5E2%2B2j-783%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']we type this quadratic equation into the search engine[/URL] and get: j = 27, j = -29. Since ages cannot be negative, we have: [B]j = 27[/B]
Sum of a number and it's reciprocal is 6. What is the number? Let the number be n. The reciprocal is 1/n. The word [I]is[/I] means an equation, so we set n + 1/n equal to 6 n + 1/n = 6 Multiply each side by n to remove the fraction: n^2 + 1 = 6n Subtract 6n from each side: [B]n^2 - 6n + 1 = 0 [/B]<-- This is our algebraic expression If the problem asks you to solve for n, then you [URL='https://www.mathcelebrity.com/quadratic.php?num=n%5E2-6n%2B1%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']type this quadratic equation into our search engine[/URL].
The difference between two positive numbers is 5 and the square of their sum is 169. Let the two positive numbers be a and b. We have the following equations: [LIST=1] [*]a - b = 5 [*](a + b)^2 = 169 [*]Rearrange (1) by adding b to each side. We have a = b + 5 [/LIST] Now substitute (3) into (2): (b + 5 + b)^2 = 169 (2b + 5)^2 = 169 [URL='https://www.mathcelebrity.com/community/forums/calculator-requests.7/create-thread']Run (2b + 5)^2 through our search engine[/URL], and you get: 4b^2 + 20b + 25 Set this equal to 169 above: 4b^2 + 20b + 25 = 169 [URL='https://www.mathcelebrity.com/quadratic.php?num=4b%5E2%2B20b%2B25%3D169&pl=Solve+Quadratic+Equation&hintnum=+0']Run that quadratic equation in our search engine[/URL], and you get: b = (-9, 4) But the problem asks for [I]positive[/I] numbers. So [B]b = 4[/B] is one of our solutions. Substitute b = 4 into equation (1) above, and we get: a - [I]b[/I] = 5 [URL='https://www.mathcelebrity.com/1unk.php?num=a-4%3D5&pl=Solve']a - 4 = 5[/URL] [B]a = 9 [/B] Therefore, we have [B](a, b) = (9, 4)[/B]
The dimensions of a rectangle are 30 cm and 18 cm. When its length decreased by x cm and its width is increased by x cm, its area is increased by 35 sq. cm. a. Express the new length and the new width in terms of x. b. Express the new area of the rectangle in terms of x. c. Find the value of x. Calculate the current area. Using our [URL='https://www.mathcelebrity.com/rectangle.php?l=30&w=18&a=&p=&pl=Calculate+Rectangle']rectangle calculator with length = 30 and width = 18[/URL], we get: A = 540 a) Decrease length by x and increase width by x, and we get: [LIST] [*]length = [B]30 - x[/B] [*]width = [B]18 + x[/B] [/LIST] b) Our new area using the lw = A formula is: (30 - x)(18 + x) = 540 + 35 Multiplying through and simplifying, we get: 540 - 18x + 30x - x^2 = 575 [B]-x^2 + 12x + 540 = 575[/B] c) We have a quadratic equation. To solve this, [URL='https://www.mathcelebrity.com/quadratic.php?num=-x%5E2%2B12x%2B540%3D575&pl=Solve+Quadratic+Equation&hintnum=+0']we type it in our search engine, choose solve[/URL], and we get: [B]x = 5 or x = 7[/B] Trying x = 5, we get: A = (30 - 5)(18 + 5) A = 25 * 23 A = 575 Now let's try x = 7: A = (30 - 7)(18 + 7) A = 23 * 25 A = 575 They both check out. So we can have
The length of a wooden frame is 1 foot longer than its width and its area is equal to 12ft The frame is a rectangle. The area of a rectangle is A = lw. So were given: [LIST=1] [*]l = w + 1 [*]lw = 12 [/LIST] Substitute equation (1) into equation (2) for l: (w + 1) * w = 12 Multiply through and simplify: w^2 + w = 12 We have a quadratic equation. To solve for w, we type this equation into our search engine and we get two solutions: w = 3 w = -4 Since width cannot be negative, we choose the positive result and have: w = [B]3[/B] To solve for length, we plug w = 3 into equation (1) above and get: l = 3 + 1 l = [B]4[/B]
The perpendicular height of a right-angled triangle is 70 mm longer than the base. Find the perimeter of the triangle if its area is 3000. [LIST] [*]h = b + 70 [*]A = 1/2bh = 3000 [/LIST] Substitute the height equation into the area equation 1/2b(b + 70) = 3000 Multiply each side by 2 b^2 + 70b = 6000 Subtract 6000 from each side: b^2 + 70b - 6000 = 0 Using our [URL='http://www.mathcelebrity.com/quadratic.php?num=b%5E2%2B70b-6000%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']quadratic calculator[/URL], we get: b = 50 and b = -120 Since the base cannot be negative, we use b = 50. If b = 50, then h = 50 + 70 = 120 The perimeter is b + h + hypotenuse Using the [URL='http://www.mathcelebrity.com/righttriangle.php?angle_a=&a=70&angle_b=&b=50&c=&pl=Calculate+Right+Triangle']right-triangle calculator[/URL], we get hypotenuse = 86.02 Adding up all 3 for the perimeter: 50 + 70 + 86.02 = [B]206.02[/B]
The product of two consecutive integers is greater than 100 Take an integer x. Next consecutive integer is x + 1 The product of those integers is: x(x + 1) This product is greater than 100 which gives us the algebraic expression of: x(x + 1) > 100 IF we want to solve for x: x^2 + x > 100 Subtract 100 from each side: x^2 + x - 100 > 0 [URL='https://www.mathcelebrity.com/quadratic.php?num=x%5E2%2Bx-100%3E0&pl=All&hintnum=+0']Solve this quadratic:[/URL]
The product of two positive numbers is 96. Determine the two numbers if one is 4 more than the other. Let the 2 numbers be x and y. We have: [LIST=1] [*]xy = 96 [*]x = y - 4 [/LIST] [U]Substitute (2) into (1)[/U] (y - 4)y = 96 y^2 - 4y = 96 [U]Subtract 96 from both sides:[/U] y^2 - 4y - 96 = 0 [U]Factoring using our quadratic calculator, we get:[/U] (y - 12)(y + 8) So y = 12 and y = -8. Since the problem states positive numbers, we use [B]y = 12[/B]. Substituting y = 12 into (2), we get: x = 12 - 4 [B]x = 8[/B] [B]We have (x, y) = (8, 12)[/B]
The Square of a positive integer is equal to the sum of the integer and 12. Find the integer Let the integer be x. [LIST] [*]The sum of the integer and 12 is written as x + 12. [*]The square of a positive integer is written as x^2. [/LIST] We set these equal to each other: x^2 = x + 12 Subtract x + 12 from each side: x^2 - x - 12 = 0 We have a quadratic function. [URL='https://www.mathcelebrity.com/quadratic.php?num=x%5E2-x-12%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']Run it through our search engine[/URL] and we get x = 3 and x = -4. The problem asks for a positive integer, so we have [B]x = 3[/B]
The square of a positive integer minus twice its consecutive integer is equal to 22. Find the integers. Let x = the original positive integer. We have: [LIST] [*]Consecutive integer is x + 1 [*]x^2 - 2(x + 1) = 22 [/LIST] Multiply through: x^2 - 2x - 2 = 22 Subtract 22 from each side: x^2 - 2x - 24 = 0 Using our [URL='http://www.mathcelebrity.com/quadratic.php?num=x%5E2-2x-24%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']quadratic calculator[/URL], we get: x = 6 and x = -4 Since the problem states [U]positive integers[/U], we use: x = 6 and x + 1 = 7 [B](6, 7)[/B]
Write this out, let the number be x. sqrt(2x) = x - 4 since 4 less means subtract Square each side: sqrt(2x)^2 = (x - 4)^2 2x = x^2 - 8x + 16 Subtract 2x from both sides x^2 - 10x + 16 = 0 Using the [URL='http://www.mathcelebrity.com/quadratic.php?num=x%5E2+-+10x+%2B+16+%3D+0&pl=Solve+Quadratic+Equation&hintnum=0']quadratic calculator[/URL], we get two potential solutions x = (2, 8) Well, 2 does not work, since sqrt(2*2) = 2 which is not 4 less than 2 However, 8 does work: sqrt(2*8) = sqrt(16) = 4, which is 4 less than the number 8.
The sum of a number and its square is 72. find the numbers? Let the number be n. We have: n^2 + n = 72 Subtract 72 from each side: n^2 + n - 72 = 0 Using our [URL='http://www.mathcelebrity.com/quadratic.php?num=n%5E2%2Bn-72%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']quadratic calculator[/URL], we have: [B]n = 8 or n = -9 [/B] Since the numbers do not state positive or negative, these are the two solutions.
The sum of the squares of two consecutive positive integers is 61. Find these two numbers. Let the 2 consecutive integers be x and x + 1. We have: x^2 + (x + 1)^2 = 61 Simplify: x^2 + x^2 + 2x + 1 = 61 2x^2 + 2x + 1 = 61 Subtract 61 from each side: 2x^2 + 2x - 60 = 0 Divide each side by 2 x^2 + x - 30 Using our [URL='http://www.mathcelebrity.com/quadratic.php?num=x%5E2%2Bx-30&pl=Solve+Quadratic+Equation&hintnum=+0']quadratic equation calculator[/URL], we get: x = 5 and x = -6 The question asks for [I]positive integers[/I], so we use [B]x = 5. [/B]This means the other number is [B]6[/B].
When 28 is subtracted from the square of a number, the result is 3 times the number. Find the negative solution. Let the number be n. Square of a number: n^2 28 is subtracted from the square of a number: n^2 - 28 3 times the number: 3n [I]The result is[/I] mean an equation, so we set n^2 - 28 = 3n n^2 - 28 = 3n Subtract 3n from each side: n^2 - 3n - 28 = 3n - 3n The right side cancels to 0, so we have: n^2 - 3n - 28 = 0 This is a quadratic equation in standard form, so we [URL='https://www.mathcelebrity.com/quadratic.php?num=n%5E2-3n-28%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']use our quadratic calculator[/URL] to solve: We get two solutions for n: n = (-4, 7) The question asks for the negative solution, so our answer is: [B]n = -4[/B]
When 4 is subtracted from the square of a number, the result is 3 times the number. Find the positive solution. Let the number be n. We have: n^2 - 4 = 3n Subtract 3n from each side: n^2 - 3n - 4 = 0 [URL='https://www.mathcelebrity.com/quadratic.php?num=n%5E2-3n-4%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']Typing this quadratic equation into the search engine[/URL], we get: n = (-1, 4) The problem asks for the positive solution, so we get [B]n = 4[/B].
Write a model that utilizes all three explanatory variables with no interaction or quadratic terms. Choose the correct answer below. A. y i = B0 + B1x1 + B2x2 + B3x3 + e i B. y i = B0 + B1x1 + B2x2 + B3x3x2 + e i C. y i = B1x1 + B2x2 + B3x3 + ei D. None of the above equations satisfy all of the conditions [B]A. y i = B0 + B1x1 + B2x2 + B3x3 + e i[/B]
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