Quadratic formula, roots

Quadratic formula, roots, 43 Quadricyclene, 203 photochemistry, 202-203 QCISD(T), 30... 

The quadratic formula yields the solution for [A] only one root is pertinent,... 

Substituting the appropriate values for K and the concentrations yields two roots of -0.0445 and 1.454. We throw out the larger root because it is a nonsensical root. This larger root wants to react more CO and H2O than are initially present. When using the quadratic formula, the user will throw out all roots greater than 1 or less than -1. The remaining root of -0.0445 is precisely what was developed by our previous trial and error exercise. 

The two roots of equation (3.15) may be abstracted by routine application of the quadratic formula. Before we do that, let us examine the results from three simplified special... 

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. 

Now the guessed value and the calculated value are the same we have found the correct solution. Note that this agrees with one of the roots found with the quadratic formula in the first method above. 

Before giving a proof, note that the components of are readily obtainable Xc-Xi and y,. is the root of a quadratic. The condition is stated in the form (3.4) to avoid the complicated expression that would result from using 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. 

Quadratic formula The roots of a polynomial of degree n = 2 are given by the quadratic formula... 

We could solve this equation using the quadratic formula, but it is easier to notice that the left side is a perfect square, so we can take the square root of both sides, giving ... 

I EXERCISE 3.1 Show by substituting Eq. (3.6) into Eq. (3.4) that the quadratic formula provides the roots to a quadratic equation. Q... 

This equation must be solved by the quadratic formula because is too large to neglect x relative to 0.100 M. Solving gives positive root x = 2.4 X 10-. Thus, from the first step in the ionization of H3PO4,... 

In general, Eq. (7=33) will be of order C in (p where C is the number of conponents. Saturated liquid and saturated vapor feeds are special cases and, after sinplification, are of order C-1. If the resulting equation is quadratic, the quadratic formula can be used to find the roots. Otherwise, a root-finding method should be employed. If only one root, aLK-ref > 9 > %K-ref> is desired, a good first guess is to... 

This equation has only one unknown, so we should be able to solve it for x and then use that result to obtain values for the final concentrations. We could multiply the denominator out and solve using the quadratic formula. Or we might save a fitde algebra by realizing that the left-hand side is a perfect square. This lets us take the square root of both sides to get a linear equation, which we can solve more easily. 

Setting b + 2x = Q m eq. (7.17), yields our bifurcation condition, — 4 = 0, or jc = 2, which corresponds to = 4. This result is, of course, easily obtained from the amnesia-blocked quadratic formula, which gives the discriminant of eq. (7.13) as b - 16. When b > 4, the discriminant is positive, and there are two real roots. When b = 4, the two roots merge, that is, the equation can be written as x 2f = 0, If Z7 <4, the discriminant is negative and there are no real roots. In other words, for b > 4, there are two real roots at the bifurcation points, b = 4, there is a single real root, x = f2 and between the bifurcation points, no real roots exist. 

In Table 7.1, we compare the exact roots of eq. (7.13) calculated from the quadratic formula with the values obtained by repeated application of eqs. (7.15) and (7.18) with a AZ> of 0.1. Even better results can be obtained by using a smaller increment in the parameter. 

Solving with the quadratic formula and taking only the positive root gives... 

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