The Quadratic Formula - Online Tutor, Practice Problems & Exam Prep

The quadratic formula, expressed as $x=-b\pm \sqrt{b²-4ac}/2a$, is essential for solving equations of the form $\mathrm{ax}²+\mathrm{bx}+c=0$. It is particularly useful in chemical equilibrium problems, where ICE charts help determine equilibrium concentrations. Typically, only one of the two potential solutions is significant, often the positive value, which will be discussed further in the context of chemical equilibrium.

Quadratic Formula

So here we are dealing with the quadratic formula. We are going to say the quadratic formula is used for algebraic equations like x2+bx+c. Okay? So here, a, and b are just numerical values. So they're actual numbers; the same thing with c as well. x are just missing variables. We're going to see the quadratic formula, which is shown here; it's x=−b±b2−4ac2a. Here it says the quadratic formula is most commonly used for questions dealing with chemical equilibrium. When we get to the sections dealing with chemical equilibrium, we are going to use things that are called ICE charts. And these ICE charts help to find equilibrium concentrations. So equilibrium amounts or concentrations, that's when we typically use these. Now you may use the quadratic formula before then within your lab. So it's still important to remember. So we're going to say although the presence of the positive negative sign gives 2 possible values for x, only one of them will be significant and used as the answer. So again, because there's a plus or minus here, that means we have a possibility of 2 answers for our x. When we do different types of questions it usually means that we have only one possibility when it comes to our final answer. Now knowing this, let's see if we can solve this question here. It says, using the quadratic formula, solve for x when given the following algebraic expression. So let me take myself out of the image guys because we need room to do this. So we're going to say here we have 4x=−2.13×10−4+1.75×10−5x. Again we want our expression to look something like this. And we want at least to have an x2 to be able to use the quadratic formula. Here we don't have that. And we have this number over x. So what we're going to do here is we're going to multiply both sides by x. And when we do that we are going to get 4x2=−2.13×10−4x+1.75×10−5. This x will cancel out with this x, so we're going to get 1.75 plus 1.75×10−5. Alright. Now we are going to say that this x variable has the larger power so it's our lead term. That means that everything has to be moved over to its side. So we are going to add 2.13×10−4x to both both sides. Then we're going to subtract 1.75×10−5 from both sides. So here what we're going to get at the end for our expression is 4x2+2.13×10−4x−1.75×10−5. So this represents my a, my b, and my c. So my quadratic formula is to remember −b±b2−4ac, divided by 2a. Alright, so here the sign of b is positive, so when I throw it in it's going to become negative now. Plus or minus b2, −4 for a, the value in front of a is a 4, and then c, the value is this, do not forget the negative sign. Okay? So if there was a negative sign here for b, let's say that this was negative, then this would have been positive, would have been the opposite, and then we would have had to put the negative there as well. Okay? So don't forget the signs get introduced as well. Divided by 2×4. Alright. So oops. So now we're going to say that I'm going to take the square root of everything in here. Okay. So what I'm going to do first, I'm going to figure out what everything is in there. Now if you do it correctly in your calculator you should get square root of 2.80×10−4 divided by 8. Then, realize you are going to get 2x variables because remember it is plus or minus. So x here equals −2.13×10−4 plus so when I take the square root of that number it is going to give me 0.016735 which gets divided by 8. So that'll give me 0.002065 for x. And then remember we can also have x=−2.13×10−4−, because remember b2, divided by 8. And in this case, x equals negative.002119. So these are my two answers that x could be. Again, when we finally use this, within our lectures, we're going to be using it for usually equilibrium questions, so chemical equilibrium questions. And only one of these answers will be the vital answer. And usually, it's the positive answer is the one that we use. We discard the negative one. But we'll talk about that later on when we're dealing with chemical equilibrium, ice charts, and all that stuff. But for now just remember what the quadratic formula bx plus bx plus c to give us the variables that we need to solve for x. We will get 2 answers and usually only one of them is a viable answer. The other one will be discarded. We will talk about the conditions and reasons for discarding which one later on.

Here’s what students ask on this topic:

What is the quadratic formula and how is it used?

The quadratic formula is used to solve quadratic equations of the form $a^{2}x+bx+c=0$. The formula is expressed as $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$. It is particularly useful in finding the roots of the equation, which are the values of x that satisfy the equation. In chemical equilibrium problems, it helps determine equilibrium concentrations using ICE charts.

When do you use the quadratic formula in chemical equilibrium problems?

The quadratic formula is used in chemical equilibrium problems when you need to solve for equilibrium concentrations. These problems often involve setting up an ICE chart (Initial, Change, Equilibrium) to organize the concentrations of reactants and products. When the equilibrium expression results in a quadratic equation of the form $a^{2}x+bx+c=0$, the quadratic formula can be applied to find the values of x. Typically, only one of the two possible solutions is physically meaningful, often the positive value.

How do you determine which solution to use from the quadratic formula?

When using the quadratic formula, $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, you get two potential solutions for x. In the context of chemical equilibrium, only one solution is typically significant. This is usually the positive value, as negative concentrations are not physically meaningful. The context of the problem and the physical constraints will guide you in selecting the appropriate solution.

What are ICE charts and how do they relate to the quadratic formula?

ICE charts (Initial, Change, Equilibrium) are used in chemical equilibrium problems to organize and calculate the concentrations of reactants and products at equilibrium. They help set up the equilibrium expression, which can sometimes result in a quadratic equation of the form $a^{2}x+bx+c=0$. The quadratic formula, $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, is then used to solve for the equilibrium concentrations.

Can the quadratic formula have more than one solution?

Yes, the quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ provides two potential solutions for x due to the ± symbol. These solutions are the roots of the quadratic equation $a^{2}x+bx+c=0$. However, in practical applications like chemical equilibrium, only one solution is often significant, usually the positive value, as negative concentrations are not physically meaningful.