Quadratic equation formula

Given the masses and force constants we can calculate A (using the quadratic equation formula) and then evaluate v = 1303A. Because Eq. (4.11) is quadratic we get two A and v values. 

This is a quadratic equation. It could be rearranged to the form ax2 + bx + c = 0 and solved for x, using the quadratic formula. Such a procedure is time-consuming and, in this case, unnecessary. Nicotinic acid is a weak acid, only slightly ionized in water. The equilibrium concentration of HNic, 0.10 — x, is probably only very slightly less than its original concentration, 0.10 M. So let s make the approximation 0.10 — x 0.10. This simplifies the equation written above ... 

Strategy The setup is identical with that in Example 13.7. However, you will find, on solving for x, that x > 0.050a, so the approximation a — x a fails. The simplest way to proceed is to use the calculated value of x to obtain a better estimate of [HNOJ, then solve again for [H+], An alternative is to use the quadratic formula. (This is a particularly shrewd choice if you have a calculator that can be programmed to solve quadratic equations.)... 

We quote below the results of computations for problem (3) with j/q = 1 and j/j =82, where is the smallest root to the quadratic equation (4). Once supplemented with those initial conditions, the exact solution of problem (3) takes the form j/, = i s (A = 0). Because of rounding errors, the first summand emerged in formula (5). This member increases along with increasing i, thus causing abnormal termination in computational procedures. 

In the proposed theory, short-range ion-solvent interaction energies are formulated as Eq. (31). the coefficients A, B, and C in the quadratic equation are related to coefficients Fj, F2, and F3 in the formula for [Eq. (36)]. The above-mentioned regression... 

As we recall from the previous chapter [5], the nonlinearity measure we devised equals the first divided by the second. Let us now develop the formula for this. We will use a subscripted small a for the coefficients of the quadratic equation, and a subscripted small k for those of the linear equation. Thus the equation describing the quadratic function fitted to the data is... 

Not all quadratic equations can be solved by factoring. And sometimes those that can be solved by factoring are more easily solved using the quadratic formula. On the other hand, all quadratic equations can be solved using the formula. It s just that factoring is usually quicker, easier, and more accurate (not as many opportunities for error). 

The quadratic formula says that if a quadratic equation is written in the form... 

One more named formula — one that all algebra students will recognize quickly — is the quadratic formula. The quadratic formula is used to find the value of the unknown variable in a quadratic equation that s written in the... 

A quadratic equation can result in two different answers. Using the quadratic formula, the two answers are found by doing the addition and subtraction in the numerator and dividing by the denominator. 

The distance formula s = -16/2 + v0t + sa gives you the height of an object when you know how much time has elapsed. You determine the input, the amount of time, by solving the quadratic equation for t. 

When solving a quadratic equation of the form ax2 + bx + c = 0, you either factor the expression and set the factors equal to 0 to solve for x, or you use the quadratic formula. (You ll find the quadratic formula in the Cheat Sheet.)... 

You end up with a quadratic equation. You can factor the equation or you can use the quadratic formula. (See the Cheat Sheet for the formula.) I show you factoring, in the following equation, after multiplying every term by 2 so that the coefficient of the x2 term becomes a 1. 

The quadratic equation is solved by either factoring or using the quadratic formula. I choose to factor, but I ll multiply each term by 2.5, first, to make the coefficient of the x2 term equal to 1. 

The quadratic equation can be factored, but the factors aren t easy to come by. Using the quadratic formula, while messy, is quicker in this case. You get that x = 24 or x is a number greater than 52. The second number doesn t make any sense and is extraneous, because the total distance down the coast is only 32 miles. Katie landed 24 miles down the coast. 

After the resulting quadratic equation is solved for [HO] and a solution of negative sign is rejected, a formula for the minimum detectable HO con-... 

In cases where factorization proves difficult, it is always possible to use the formula for the roots of a quadratic equation, ax + bx + c=0 ... 

A quadratic equation is an equation in which the greatest exponent of the variable is 2, as in x2 + 2x - 15 = 0. A quadratic equation had two roots, which can be found by breaking down the quadratic equation into two simple equations. You can do this by factoring or by using the quadratic formula to find the roots. 

Solving Quadratic Equations Using the Quadratic Formula... 

The standard form of a quadratic equation is ax2 + bx + c = 0, where a, b, and c are real numbers (a = 0). To use the quadratic formula to solve a quadratic equation, first put the equation into standard form and identify a, b, and c. Then substitute those values into the formula ... 

Accordingly, the solution of (74) was found from a quadratic equation. It is clear that the possibilities to produce exact explicit formulas along this line are limited the solutions for n = 4 and n = 5 have been accomplished [17, 18] by means of the relevant cubic equations, but the answer for larger n values have not been found previously. 

Let us use a process similar to finding solutions to ordinary simultaneous equations, in which variables arc changed by taking linear combinations such that the new variables form sets that are not coupled to each other. Then, smaller sets of equations can be solved, and finally the equations can be reduced to coupled pairs that can be solved by using the formula for the roots of a quadratic equation. 

Note that although we used the quadratic formula to solve for x in this problem, other methods are also available. For example, trial and error is always a possibility. However, use of successive approximations (see Appendix A1.4) is often preferable. For example, in this case successive approximations can be carried out conveniently by starting with the quadratic equation... 

This expression produces a quadratic equation that can be solved by using the quadratic formula, as for the gas-phase systems in Chapter 6. However, since Ka for HF is so small, HF will dissociate only slightly thus x is expected to be small. This will allow us to simplify the calculation. If x is very small... 

One method for finding the two values of x that satisfy a quadratic equation is to use the quadratic formula ... 

Sometimes a chemical problem can be reduced algebraically, by pencil and paper, to a polynomial expression for which the solution to the problem is one of the roots of the polynomial. Almost everyone remembers the quadratic formula for the roots of a quadratic equation, but finding the roots of a more complicated polynomial is more difficult. We begin by describing three methods for finding the real roots of a polynomial. 

This quadratic equation can be solved by use of the quadratic formula, provided the calculator carries ten significant figures. It can be solved more easily by making the approximation that y is much smaller than 0.0015 the equation therefore simplifies to... 

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