ROOTS of quadratic equation help?
This is a great problem if there's an elegant solution. I'll keep an eye out to see if somebody finds one. In the meantime, here's an ugly brute-force solution. First of all, I need clarification. Is α²+β² one root, or are α² and β² the two roots? My answer assumes the second option since the question suggests that we are being given both roots. Now continuing on. It's easy to find the two complex roots by the quadratic equation. Set one equal to α² and the other equal to β². (i) They don't say that the polynomial must have real coefficients, so let p(x) = (x - (1/α²))*(x - (1/β²)). That's a quadratic with roots 1/α² and 1/β². (ii) Let q(x) = (x-α)(x-β). Then q(x) and 2*q(x) are two distinct quadratics with roots α and β. Yes it's a cheap solution, but it fits all the criteria as specified in the problem.
Quadratic Equation help!?
Under what conditiond would the well known quadratic formula not be effectively computable? (Assume that you are working with real numbers.) If you could provide explanation so I am able to have a better understanding.
Quadratic Equation help me?
The standard form of a quadratic equation is ax^2 + bx + c = 0 What you have is not correct. It can be a quadratic function (not a standard form though). In these two problems: remove the brackets: 3(x^2 + 14x + 49) + 4 = 79 3x^2 + 42x + 147 + 4 = 79 now to the standard form: 3x^2 +42x + 72 = 0 divide both sides by 3 x^2 + 14x + 24 = 0 factorize: (x + 12)(x + 2) = 0 x = -12 or x = -2 follow the same steps for 2 Good Luck, Kempos
Why do we use quadratic equations?
There are many properties of the world that behave exactly or nearly like a parabola, such as an arc of water coming out of a spout or hose, or the trajectory of a ball or other missile in the air.
Quadratic Equations homework help PLEASE?
Find the first derivative of the function C = 0.000015X^2 - 0.03x + 35: dC/dX = 2*0.000015X - 0.03 Equate dC/dX to 0, then solve for X: dC/dX = 2*0.000015X - 0.03 = 0 X = 0.03 / (2*0.000015) X = 1000 tires <=== number of tires to minimize cost C = 0.000015X^2 - 0.03x + 35 C = 0.000015(1000)^2 - 0.03(1000) + 35 C = $20 <=== the cost of producing 1000 tires; seems unrealistic to me.
I need math help! I need to find a quadratic equation from this word problem.?
Sounds like y = (x^2)/72
How do you understand quadratic equations?
One practical example of a quadratic equation (there are many) is finding the stopping distance for a car or any vehicle. The stopping distance is made up of the distance that is traveled during the driver’s reaction time and then once the brakes are applied, the distance that is travelled during the period of deceleration. In the general model,ax^2 + bx + c =0,the bx component relates to the thinking distance, which is linear. The ax^2 component is non-linear, because if the speed is doubled from v to 2v, then thestopping distance is quadrupled.If you go to web sitehttp://passmytheory.co.uk/learni...you will see various thinking and braking distances giving overall stopping distances. Modeling this site’s figures gives the following quadratic, which will yield total stopping distances for any normal speed of a car:x^2/20 + x - c = 0where x is the speed of the car in mph and c is the overall stopping distance in feet.Suppose we know the stopping distance as 75 feet (= c), find the speed (x mph) at which the vehicle was traveling? The model isx^2/20 + x - 75 = 0Do the calculation > x^2 + 20x -1500 =0 using the formula and ignore the negative result. This yields that x = 30 (mph)Alternatively, you may have the speed of the car (say it is 20 mph so x = 20) and require the stopping distance (= c) and so the formula becomes:x^2/20 + x = c => 20^2/20 + 20 = 40 (feet)I hope this is a useful example to help the understanding of quadratic equations.