In this video, we will learn how to solve quadratic equations using the quadratic formula.

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### Video Transcript

In this video, we will learn how to solve quadratic equations using the quadratic formula. The quadratic formula in its current form was first recorded in the 17th century by the French mathematician Rene Descartes. It is an efficient way to solve a quadratic equation, especially one which cannot be solved by factoring. Given any quadratic equation in the form 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero, where 𝑎, 𝑏, and 𝑐 are constants and 𝑎 is nonzero, then the quadratic formula states that 𝑥 is equal to negative 𝑏 plus or minus the square root of 𝑏 squared minus four 𝑎𝑐 all divided by two 𝑎. Let’s now consider how we can use this formula to solve any problem involving a quadratic equation.

Our first step is to write our equation in the form 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero. In order to do this, we might need to distribute parentheses or brackets as well as collect like terms. Whilst it is not essential, it is useful to write our equation in the same order. It is then easy to list the values of 𝑎, 𝑏, and 𝑐. Next, we substitute our values of 𝑎, 𝑏, and 𝑐 into the quadratic formula. From this point, we simplify the right-hand side obtaining two solutions to the quadratic equation. Let’s consider the quadratic equation three 𝑥 squared plus four 𝑥 minus one is equal to zero. Our values of 𝑎, 𝑏, and 𝑐 are three, four, and negative one, respectively, as the coefficient of 𝑥 squared is three, the coefficient of 𝑥 is four, and the term independent of 𝑥 is negative one.

Substituting these values into the quadratic formula gives us 𝑥 is equal to negative four plus or minus the square root of four squared minus four multiplied by three multiplied by negative one all divided by two multiplied by three. Underneath the square root, our expression simplifies to 16 plus 12. This gives us negative four plus or minus the square root of 28 all divided by six. The square root of 28 is equal to two root seven. We can then divide each term in our expression by two. 𝑥 is therefore equal to negative two plus or minus root seven all divided by three. We can separate this to get the two solutions to the quadratic equation three 𝑥 squared plus four 𝑥 minus one equals zero. Either 𝑥 is equal to negative two plus root seven divided by three or 𝑥 is equal to negative two minus root seven divided by three.

We will now look at a more complicated example.

Find the solution set of the equation three 𝑥 squared plus four multiplied by 𝑥 plus one equals zero, giving values in the set of real numbers to one decimal place.

In order to answer this question, we firstly need to rearrange our equation so it is in the form 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero. This will enable us to use the quadratic formula to solve it. Distributing the parentheses by multiplying four by 𝑥 and four by one gives us three 𝑥 squared plus four 𝑥 plus four equals zero. Our values of 𝑎, 𝑏, and 𝑐 are three, four, and four, respectively. The quadratic formula states that 𝑥 is equal to negative 𝑏 plus or minus the square root of 𝑏 squared minus four 𝑎𝑐 all divided by two 𝑎.

Substituting in our values, we have 𝑥 is equal to negative four plus or minus the square root of four squared minus four multiplied by three multiplied by four all divided by two multiplied by three. Four squared is 16, four multiplied by three multiplied by four is 48, and two multiplied by three is six. As 16 minus 48 is negative 32, we are left with 𝑥 is equal to negative four plus or minus the square root of negative 32 all divided by six.

At this stage, as we want solutions to one decimal place, we would usually input our calculation into our calculator. However, this calculation involves taking the square root of a negative number, negative 32. And we know that square rooting a negative number has no real solutions. And when we type it into the calculator, we get a mathematical error. This means that there are no real solutions to our equation. And the solution set of the equation is the empty set. This leads us to a key fact about the quadratic formula. If the expression underneath the square root 𝑏 squared minus four 𝑎𝑐, known as the discriminant, is less than zero, then there are no real solutions to our quadratic equation.

It is also worth considering what this would look like graphically. The quadratic equation 𝑦 is equal to three 𝑥 squared plus four multiplied by 𝑥 plus one is shown in the figure. We notice that the graph does not intersect the 𝑥-axis. This confirms that there are no real solutions to the equation. Any quadratic equation, where the discriminant 𝑏 squared minus four 𝑎𝑐 is less than zero, will not intersect the 𝑥-axis.

We will now look at two further examples where we need to use the quadratic formula in context.

The dimensions of a rectangle are five meters and 12 meters. When both dimensions are increased by a given amount, the area of the rectangle will double. What is the amount?

We are given that the dimensions of a rectangle are five meters and 12 meters. Recalling that we can calculate the area of a rectangle by multiplying its length by its width, the area of this rectangle is 60 square meters. Next, we are told that both dimensions are increased by a given amount. We will let this given amount be 𝑥 meters so that the length of the new rectangle is 12 plus 𝑥 meters and the width is five plus 𝑥 meters.

To calculate the area of this rectangle, we need to multiply 12 plus 𝑥 and five plus 𝑥. We are also told that the area of the rectangle has doubled. Therefore, the area of this rectangle is 120 square meters. Distributing our parentheses using the FOIL method, we have 60 plus 12𝑥 plus five 𝑥 plus 𝑥 squared. And this is equal to 120. Subtracting 120 from both sides of this equation and then collecting like terms, we have 𝑥 squared plus 17𝑥 minus 60 is equal to zero. This is a quadratic equation written in the form 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero. And we know we can solve a quadratic equation of this type using the quadratic formula. This states that 𝑥 is equal to negative 𝑏 plus or minus the square root of 𝑏 squared minus four 𝑎𝑐 all divided by two 𝑎.

In this question, our values of 𝑎, 𝑏, and 𝑐 are one, 17, and negative 60, respectively. Substituting these values into the formula, we have 𝑥 is equal to negative 17 plus or minus the square root of 17 squared minus four multiplied by one multiplied by negative 60 all divided by two multiplied by one. The expression under the square root is equal to 529. And as the square root of 529 is 23, we have 𝑥 is equal to negative 17 plus or minus 23 all divided by two. This gives us two possible solutions, either 𝑥 is equal to negative 17 plus 23 divided by two or 𝑥 is equal to negative 17 minus 23 all divided by two. These are equal to three and negative 20, respectively.

Since we are solving to find the dimension 𝑥 of the rectangle, we know that 𝑥 must be positive. This means that the correct solution is 𝑥 equals three. For the area of the rectangle to double, both dimensions need to be increased by 3 meters. This would give us a rectangle with dimensions eight meters by 15 meters. And multiplying these gives us the correct area of 120 square meters. It is important to note that this quadratic equation could have been factored. Using this method would also have given us solutions of 𝑥 equals negative 20 and 𝑥 equals three.

We will now consider one final example.

A stone is thrown upward from the top of a cliff and lands in the sea some time later. The height ℎ above sea level at any time 𝑡 seconds is given by the formula ℎ is equal to three 𝑡 minus five 𝑡 squared plus 40. After how many seconds will the stone reach the sea? Give your answer to the nearest tenth of a second.

In this question, we are given a quadratic equation for ℎ in terms of 𝑡. Rewriting the right-hand side in the form 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐, we have ℎ is equal to negative five 𝑡 squared plus three 𝑡 plus 40. We are trying to calculate the time at which the stone will reach the sea, and we are told that ℎ is the height above sea level. This means that when the stone reaches the sea, ℎ will be equal to zero. And we need to solve the quadratic equation negative five 𝑡 squared plus three 𝑡 plus 40 equals zero. We know that any quadratic equation 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero can be solved using the quadratic formula. This states that 𝑥 is equal to negative 𝑏 plus or minus the square root of 𝑏 squared minus four 𝑎𝑐 all divided by two 𝑎.

In this question, our values of 𝑎, 𝑏, and 𝑐 are negative five, three, and 40, and the variable instead of being 𝑥 is the time 𝑡. Substituting in our values, we have 𝑡 is equal to negative three plus or minus the square root of three squared minus four multiplied by negative five multiplied by 40 all divided by two multiplied by negative five. This simplifies to negative three plus or minus the square root of 809 all divided by negative 10. Separating our two solutions, we have 𝑡 is equal to negative three plus the square root of 809 divided by negative 10 or 𝑡 is equal to negative three minus the square root of 809 divided by negative 10.

Typing these into the calculator, we have 𝑡 is equal to negative 2.544 and so on and 𝑡 is equal to 3.144 and so on. As the time in seconds must be a positive value, we can rule out the first answer. We are also asked to give the time to the nearest tenth of a second. We can therefore conclude that it takes 3.1 seconds for the stone to reach the sea.

We will now summarize the key points from this video. For any quadratic equation in the form 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero, where 𝑎, 𝑏, and 𝑐 are constants and 𝑎 is nonzero, we can use the quadratic formula to solve for 𝑥. This states that 𝑥 is equal to negative 𝑏 plus or minus the square root of 𝑏 squared minus four 𝑎𝑐 all divided by two 𝑎. The solutions to the equation are the points where the graph of the equation crosses the 𝑥-axis. If the discriminant 𝑏 squared minus four 𝑎𝑐 is less than zero, then the quadratic equation has no real solutions. This means that the graph would not pass through the 𝑥-axis. Whilst it is not included in this video, it may be of interest to research how the quadratic formula can be derived.