The Quadratic Formula

Suppose we were given instead x + 2x + 7. We caimot factor this as readily, but here is an alternative trick. Knowing how the polynomial (3.58) factors we can write 

This suggests how to solve the quadratic equation (3.57). First, complete the square in the first two terms  

The simplest quadratic equation with imaginary roots is 

Observe that whenever D 0, the roots occur as conjugate pairs, one root containing — D and the other containing — y — Z).  

The three quadratic equations considered above can be solved graphically (Fig. 3.3). The two points where the parabola representing the equation crosses the x axis correspond to the real roots. For a double root, the curve is tangent to the x axis. If there are no real roots, as in the case of x -f 1 = 0, the curve does not intersect the x axis. 

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H30+]2 + fCa[H30+] - fCaCHF = 0 which we solve using the quadratic formula... 

Solving the rightmost eqiiahty for the directional derivative using the quadratic formula gives... 

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 ... 

In most of the problems you will work, the approximation a — x a is valid, and you can solve for [H+] quite simply, as in Example 13.7, where x = 0.012a. Sometimes, though, you will find that the calculated [H+] is greater than 5% of the original concentration of weak acid. In that case, you can solve for x by using either the quadratic formula or the method of successive approximations. 

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.)... 

WLM liked this method CNH leans toward the use of the quadratic formula. 

The quadratic formula. This gives an exact solution for x but is more time-consuming. Rewrite the equation... 

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

As shown in Example 10.14, we can use the quadratic formula to solve this equation for the concentration of hydronium ions. 

The concentration of hydronium ions can be obtained by solving this equation with the quadratic formula. 

In cases where x is too large to neglect from the concentration from which it is subtracted, a more exact method is required. The quadratic formula is an exact method to determine the value of x in a equation of the form... 

This value of x is 42.4% of the concentration from which it was subtracted. The approximation is wrong. The quadratic formula must be used ... 

Here the a cannot be neglected when subtracted from the 10 because the equilibrium constant is too big. The problem is solved using the quadratic formula. 

Equation 66-A14 is quadratic in Z3, and thus, after evaluating the summations is easily solved through use of the Quadratic Formula. 

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. 

You ll find lots of cross-referencing in the chapters. If a problem requires the use of the quadratic formula, I send you to the chapter or section where I introduce that formula. Each section and each chapter stands by itself — you don t really need to go through the chapters sequentially. You re more than welcome to go back and forth as much as you want. This isn t a murder mystery where the whole plot will be exposed if you go to the end first. When reading this book, do it your way ... 

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... 

Use the quadratic formula, letting a = 12, b = -31, and c = 20. First, simplify what s under the radical, which acts as a grouping symbol. Do the power first, multiply the three terms together, then subtract the product from the power. 

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. 

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. 

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