real life examples of quadratic equations

Best Results From Yahoo Answers Encyclopedia


From Encyclopedia

Quadratic Formula and Equations Quadratic Formula and Equations

A quadratic equation is an equation of the second degree, meaning that for an equation in x, the greatest exponent on x is 2. Quadratics most commonly refer to vertically oriented parabolas—that is, parabolas that open upward or downward. The graph of a vertically oriented parabola has the shape of a rounded "v," and the bottom-most (or top-most) point is called the vertex. The equation for a parabola is usually written in either standard or vertex form; however, the standard form is more commonly used to solve for the x -intercepts, or roots. The standard form is y = ax 2 + bx + c for any real numbers a, b, c where a ≠ 0. The vertex form is y − k = a (x − b )2 with vertex (b, k ) and where a ≠ 0. Because x -intercepts are the points at which the graph crosses the x -axis, the solutions are always found by substituting 0 for y. The roots are often useful in solving real world problems, and there are three common ways to find the roots: factoring, using the quadratic formula, and completing the square. Not all quadratics can easily be factored, but if they can, the quickest way to solve them is to factor and use the zero product property. The zero product property basically states that if the product of two numbers is 0, then at least one of the numbers multiplied must be a 0. In other words, for any real numbers a and b, if ab = 0, then either a = 0 or b = 0. Consider a swimmer who starts at one end of a pool, swims down to pick up a ring at the bottom of the middle of the pool, and then surfaces at the other end of the pool with the ring. The equation y = x 2 − 6x + 9 can be used to model the path of the swimmer, where y is the water level in the pool measured in feet, and x is the time in seconds since the swimmer started. The equations below show how to solve for the roots of the equation to find the number of seconds it took the swimmer to reach the ring at the bottom of the pool: namely, by substituting 0 for y, factoring, and using the zero product property. 0 = x 2 − 6x + 9 Substitute 0 for y. 0 = [x − 3](x − 3) Factor x 2 − 6x + 9. Either [x − 3] = 0 or (x − 3) = 0 Use the zero product property. So x = 3 and x = 3 Solve each equation. In this example, the quadratic has only one repeated root, x = 3. This root is the time at which the swimmer reached the bottom of the pool. This quadratic can be graphed by substituting values into the equation to make a table of points, then graphing the realistic portion of the parabola, as shown below. The graph below illustrates that the parabola has a vertical line of symmetry that passes through (3, 0). The equation for the line of symmetry of this parabola is x = 3. The graph of the parabola continues infinitely; however, to model the path of the swimmer, only the points from 0 to 6 seconds are graphed. This is because the swimmer starts at x = 0 seconds, swims down for 3 seconds to get the ring, and then swims up for 3 seconds. Because not all quadratic equations can be factored, other methods for finding roots are needed. One other method of finding roots for a quadratic is to use the quadratic formula. In the formula, the plus or minus sign means to solve the formula twice—once with a plus, and once with a minus. In other words, given a quadratic equation in standard form, y = ax 2 + bx + c, the solutions can be found by Consider that a delayed space shuttle leaves Earth about 20 minutes after the scheduled departure. At 6 miles out, the shuttle turns around and returns to Earth. The distance of the shuttle from Earth can be described by the equation y − 6 = −0.1(x − 30)2, where x is the number of minutes the shuttle is in flight. The equations below show how to find the total number of minutes the shuttle was off the ground. To find the roots of the equation, first solve for standard form and substitute 0 for y, as shown in Step One. The resulting trinomial cannot easily be factored into two binomials, so the quadratic equation must be used to solve for the roots, as shown in Step Two. Step One Step Two To graph the parabola, plot and connect the two roots and the vertex. (The equation was originally given in vertex form.) If needed, more points can be found by substituting values for x into the equation. To graph the realistic portion of the parabola, graph only the portion in Quadrant I (see below). The original problem said that the shuttle was delayed by about 20 minutes. This is the first intercept, x ≈ 22.25 minutes. The vertex represents the point at which the shuttle was 6 miles from Earth. The second intercept, x ≈ 37.75 represents the time at which the shuttle returned to Earth. To find the total number of minutes the shuttle was in flight, subtract its liftoff and landing times. The shuttle was in flight for about 37.75 − 22.25 = 15.5 minutes. Some equations for parabolas may be solved more easily by completing the square. Completing the square forces the trinomial to be a perfect square by replacing the constant term with . In addition to finding roots, completing the square is also used for transforming an equation in standard form to vertex form. Furthermore, this method can be extended for use with the other conic sections . The quadratic formula can be derived by completing the square in y = ax 2 + bx + c. The equations below show how to solve for vertex form of y = x 2 – 6x + 7 and find its roots by completing the square. The strategy is to move the constant term opposite the trinomial and replace it with Then the new trinomial is written as the square of a binomial. In Step Two, the vertex form is y + 2 = (x − 3)2, and the vertex is (3 − 2). To find the roots, substitute 0 for y and solve for x. The two roots are and . To graph this parabola, the vertex and approximations for the roots can be plotted and connected. Step One Step Two b)2. Then The vertex can be found directly from vertex form, and it can also be found from standard form. From standard form, use b /2a to find the x -coordinate of the vertex and then substitute the result into the equation to find the y -coordinate of the vertex. The discriminant can be used to determine if the graph crosses the x -axis, and if so how many times. The discriminant is the expression under the radical in the quadratic formula, b 2 − 4ac. A square root usually yields two solutions, unless it is the square root of zero. The table summarizes the number and types of solutions that can occur and how they affect the appearance of the graph. The value of a in a quadratic equation also affects the placement of the graph on the plane. If a is positive, the graph opens upwards; if it is negative the graph opens downward. If a is greater than one, the graph will be narrow, and if a is a fraction between 0 and 1, the graph will be wide. This bit of information is especially useful because the value of a affects other types of graphs in the same ways as it does parabolas. All conic sections are quadratics because they have equations of the second degree. However, only the vertically oriented parabolas that have been summarized in this article are functions. Graphing calculators and computers perform functions by taking an input and giving an output. Hence, most graphing tools are only equipped to graph equations of functions. To graph a horizontally oriented parabola on a calculator, the graph must be broken into pieces that are functions. Then the equations for each piece are graphed on the same plane to create the appearance of one graph. see also Conic Sections; Functions and Equations; Graphs and Effects of Parameter Changes. Michelle R. Michael


From Yahoo Answers

Question:I am needing to have a real life quadratic equation example for school and I don't even know where to start!!! I keep finding examples that are way too complex and I don't quite understand it yet..... I need some help!

Answers:One way of approaching this is to work backwards. Find a simple quadratic, and then think of words that might represent a real-life situation. Often, these real-life situations involve circumstances where you have two unknown numbers, but you know two things about them: 1. Their sum or difference 2. Their product. To find a simple quadratic, note that, if you take two expressions involving x and multiply them, the result will be a quadratic. Go with something simple, say: (x - 2)(x + 3) = 0 x^2 + x - 6 = 0 x^2 + x = 6 Then, refactor the left-hand side to get: x(x + 1) = 6 So, now we have two numbers (x and x + 1). We know their difference is 1, and their product is 6. Pick two things to be represented by x and x + 1, respectively. Say x represents pairs of shoes, and x + 1 represents belts. Then, x(x + 1) is the number of possible different combinations that can be made from x pairs of shoes and x + 1 belts. We already know this number is 6. So, our "real-life problem" might be something like this: Sally is flying to London to conduct a six-day seminar on human rights. Due to space limitations, she must pack one fewer pairs of shoes than belts. How many shoes and belts must Sally pack, if she doesn't want to wear the same shoe-belt combination twice? Now, working forwards, we would set up the problem like this: Let x = the number of pairs of shoes Sally packs. Then, x + 1 = the number of belts Sally packs. The number of shoe-belt combinations is: x(x + 1) = 6 x^2 + x - 6 = 0 (x - 2)(x + 3) = 0 x = {-3, 2} Since Sally can't have -3 pairs of shoes, the solution is 2 pairs of shoes, which means 2 + 1 = 3 belts. OK, this isn't exactly "real life" stuff (why would anybody ever be required to pack exactly one fewer pairs of shoes than belts?) but it gives a good idea of how these word problems are put together: Work backwards from the solution, and find words to symbolize what the solution actually means.

Question:I need a real life example of a quadratic equation (ax2+bx+c=0) Also two examples for x would be great, thanks.

Answers:x2 +5x+9=0 Then plug it into the quadratic formula.

Question:I have a project where I have to write a 5 page essay about quadratic equations. I have heard about people using them for missiles in the army, and satellite dishes. If you have any information, please, PLEASE, help me.

Answers:Yes, missiles and other projectiles under the influence of gravity (or some other accelerating agent), will follow a parabolic trajectory (if we ignore air resistance and some other factors). For example, at time t, the vertical height of an object launched into the air will be: x(t) = (1/2)*g*t^2 + v0*t + x0 where t represents the time since the object was launched, g represents the acceleration due to gravity, v0 represents the objects initial vertical velocity, and x0 represents the objects initial vertical height. For more information, try searching for "projectile motion," "free-fall motion," "trajectory," etc.

Question:I have an assignment for my Algebra 1 class, I'm a freshman. Our assignment says "You will create original word problems, revelant & meaningful in your life, which can be solved using equations. You must have one word problem & solution representing each an equation with addition, subtraction, multiplication, and divison, one multi-step equation, and one equation with variables in both sides." Does anyone have an idea of a real life example that could fit any type of the above equations?

Answers:If you have 4 cats....4c and two get hit by a bus....- 2c how many cats do you have? 4c - 2c = 2c {4 cats minus 2 dead cats equals 2 cats} 2c = 2c {combined like terms on left} 0 = 0 {subtracted 2c from both sides} when the variable disappears and you get a statement which is alway true, there are "infinitely many solutions".... guess that means the cats have more than one life! By the way, tell your teacher to teach you, instead of you teaching him/her. http://www.algebrahouse.com