Lesson Explainer: Solving Quadratic Equations: Quadratic Formula Mathematics

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

Mathematicians have been working with quadratic equations in different forms for several thousand years, with the first records of these in 500 BC by Pythagoras and 300 BC by Euclid. However, it was in the 17th Century that the mathematician Rene Descartes first recorded the quadratic formula in the form we know. The quadratic formula is an efficient way to solve a quadratic equations, especially those which cannot be solved by factoring.

Definition: Quadratic Formula

If 𝑎𝑥+𝑏𝑥+𝑐=0, where 𝑎, 𝑏, and 𝑐 are constants and 𝑎≠0, then 𝑥=−𝑏±√𝑏−4𝑎𝑐2𝑎.

It is most important to know how to use this formula. However, knowing how to derive it is also of interest. We therefore, include a derivation of this formula at the end of the explainer. To use the quadratic formula to solve an equation, we substitute in the values of 𝑎, 𝑏, and 𝑐 from our equation into the quadratic formula. In the box below, we detail how to use the quadratic formula.

How To: Using the Quadratic Formula

  • We first need to have an equation in the form 𝑎𝑥+𝑏𝑥+𝑐=0. We may need to expand any brackets or transform the terms to obtain this. It is often helpful to write the terms in our equation in the same order as in 𝑎𝑥+𝑏𝑥+𝑐=0, with the 𝑥, 𝑥, and constant terms in that order. Writing it in this form can help us ensure that we take the correct 𝑎, 𝑏, and 𝑐 values. It is also useful to list our values 𝑎=, 𝑏=, 𝑐=, so that these can be referenced when substituting.
  • We should start our working by writing the quadratic formula fully. This will help avoid errors and ensure that the positive and negative values are consistently applied throughout our working.
  • We substitute our 𝑎, 𝑏, and 𝑐 values into the quadratic formula. It is important to write any negative number substitutions in brackets as this will help maintain the accuracy of the positive and negative values as we go through the steps of the solution.

We will now look at some examples where the quadratic formula can be used to solve an equation.

Example 1: Solving an Equation Using the Quadratic Formula

Solve the equation −𝑥+7𝑥+1=0.

Answer

An equation of the form 𝑎𝑥+𝑏𝑥+𝑐=0 can be solved using the quadratic formula 𝑥=−𝑏±√𝑏−4𝑎𝑐2𝑎.

In the equation −𝑥+7𝑥+1=0, we have 𝑎=−1, 𝑏=7, and 𝑐=1. Therefore, we can substitute these values into the formula and simplify the terms, giving us 𝑥=−7±√7−4(−1)(1)2(−1).

Simplifying the terms, paying careful attention to the negative signs, we have 𝑥=−7±√49+4−2=−7±√53−2=7±√532.

Hence, we have the solution set 7−√532,7+√532

Example 2: Solving an Equation Using the Quadratic Formula

Find the solution set of the equation 3𝑥+4(𝑥+1)=0 in ℝ.

Answer

In order to begin solving this, we first expand the brackets to give us an equation in the form 𝑎𝑥+𝑏𝑥+𝑐=0: 3𝑥+4(𝑥+1)=03𝑥+4𝑥+4=0.

This equation can now be solved using the quadratic formula 𝑥=−𝑏±√𝑏−4𝑎𝑐2𝑎.

In the equation 3𝑥+4𝑥+4=0, we have the values 𝑎=3, 𝑏=4, and 𝑐=4. Therefore, we can substitute these values into the formula, giving us 𝑥=−4±√4−(4)(3)(4)2(3)=−4±√16−486=−4±√−326.

However, we can see that our calculation includes taking the square root of a negative number, √−32, which will have no real solutions. If we were to input this into a calculator, we would get a mathematical error. Therefore, there are no real solutions to 3𝑥+4(𝑥+1)=0. Hence, the solution set of the equation is the empty set ∅.

It is important to note that although the quadratic formula breaks down when the discriminant, 𝑏−4𝑎𝑐, of the calculation is less than zero, it simply means that the graph of this function would not pass through the 𝑥-axis, meaning that there are no solutions for 𝑥. If we were to draw a graph of 𝑦=3𝑥+4(𝑥+1), it would look as below.

Example 3: Finding the Roots of an Equation Using the Quadratic Formula

Given that 𝑥=−1 is one of the roots of the equation −3𝑥−9𝑥+𝑘=0, find the other root and the value of 𝑘.

Answer

We find the roots of an equation by calculating what the values of 𝑥 are when the 𝑦-value is equal to zero. When we have an equation in the form 𝑎𝑥+𝑏𝑥+𝑐=0, where 𝑎, 𝑏, and 𝑐 are constants, we can find the the values of 𝑥, the roots, by using the quadratic formula 𝑥=−𝑏±√𝑏−4𝑎𝑐2𝑎.

Although we do not yet know the value of 𝑘 in the equation −3𝑥−9𝑥+𝑘=0, we can substitute in the values 𝑎=−3, 𝑏=−9, and 𝑐=𝑘 into the quadratic formula, giving us 𝑥=9±√(−9)−(4)(−3)(𝑘)2(−3).

Simplifying our terms, we have 𝑥=9±√81+12𝑘−6.

We were given that one of the roots was −1, so we can substitute 𝑥=−1 into our equation, giving −1=9±√81+12𝑘−6.

Multiplying both sides of the equation by −6, we have 6=9±√81+12𝑘.

Subtracting 9 from both sides gives us −3=±√81+12𝑘.

Squaring both sides, we have 9=81+12𝑘.

We can then subtract 81 from both sides and simplify the terms in order to find the value of 𝑘, which gives us 9−81=12𝑘−72=12𝑘𝑘=−6.

We can substitute 𝑘=−6 into our earlier workings where 𝑥=9±√81+12𝑘−6. This gives us 𝑥=9±√81−72−6=9±√9−6=9±3−6.

So our solutions are 𝑥=−126𝑥=−66𝑥=−2𝑥=−1.oror

Therefore, the roots of the equation are 𝑥=−2 and 𝑥=−1. As we were given the root 𝑥=−1, this means that our answer is 𝑥=−2 and 𝑘=−6.

In the next example, we will see a question where it is not immediately obvious that there is a quadratic equation. It is always useful to set out the information given in a problem and write it mathematically. In the following question, the presence of an unknown, denoted 𝑥, and a multiplication leads us to forming a quadratic equation which can then be solved using the quadratic formula.

Example 4: Solving a Problem Using the Quadratic Formula

The dimensions of a rectangle are 5 m and 12 m. When both dimensions are increased by a given amount, the area of the rectangle will double. What is the amount?

Answer

The area, 𝐴, of the original rectangle can be written as 𝐴=5×12=60.

Let the amount of increase be 𝑥. We are given that the area is doubled when 𝑥 is added to the dimensions of the rectangle, so the new area would be (5+𝑥)(12+𝑥)=120.

We can expand the brackets and rearrange the equation into the form 𝑎𝑥+𝑏𝑥+𝑐=0, with constants 𝑎, 𝑏, and 𝑐, giving us 60+17𝑥+𝑥=120 which we can rewrite as 𝑥+17𝑥−60=0.

Quadratic equations in the form 𝑎𝑥+𝑏𝑥+𝑐=0 can be solved using the quadratic formula 𝑥=−𝑏±√𝑏−4𝑎𝑐2𝑎.

We can substitute the values 𝑎=1, 𝑏=17, and 𝑐=−60 into the quadratic formula and simplify, giving us 𝑥=−17±√17−4(1)(−60)2×1.

Simplifying the terms, paying careful attention to the negative signs within the square root, we have 𝑥=−17±√289+2402=−17±√5292=−17±232.

Therefore, 𝑥=−17+232𝑥=−17−232𝑥=62𝑥=−402.oror

So, 𝑥=3𝑥=−20.or

In this question, we are solving to find the dimension, 𝑥, of the rectangle. Therefore, we require a positive value of 𝑥, so 𝑥=3. The increase is 3 m.

In this question, we could have solved using an alternative method of factoring, as the rearranged equation 𝑥+17𝑥−60=0 has simple factors. We can factor the equation and solve to give 𝑥+17𝑥−60=0(𝑥+20)(𝑥−3)=0.

Therefore, 𝑥=−20𝑥=3.or

In the same way as above, we require a positive dimension and reject the negative value of 𝑥. So the required answer is 𝑥=3; therefore, the increase is 3 m.

Let us now look at an example of an equation in a real-life context, which is not a quadratic in 𝑥. We follow the same process of using the quadratic formula with the coefficients for values 𝑎, 𝑏, and 𝑐 in the same way as we do for a quadratic in 𝑥.

Example 5: Solving a Real-Life Problem Using the Quadratic Formula

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=3𝑡−5𝑡+40.

After how many seconds will the stone reach the sea? Give your answer to the nearest tenth of a second.

Answer

The equation ℎ=3𝑡−5𝑡+40 is a quadratic equation in 𝑡. We can rewrite it into the form ℎ=−5𝑡+3𝑡+40.

When the stone reaches the sea, this will mean that the height above sea level is zero, ℎ=0, so this gives us −5𝑡+3𝑡+40=0.

Recall that a quadratic equation in 𝑥, in the form 𝑎𝑥+𝑏𝑥+𝑐=0, can be solved using the quadratic formula 𝑥=−𝑏±√𝑏−4𝑎𝑐2𝑎.

Since we have the form 𝑎𝑡+𝑏𝑡+𝑐=0, we can solve with the quadratic formula written as 𝑡=−𝑏±√𝑏−4𝑎𝑐2𝑎.

Now we can substitute the values𝑎=−5, 𝑏=3, and 𝑐=40 into the quadratic formula and simplify, which gives us 𝑡=−3±√3−4(−5)(40)2(−5).

Simplifying the terms in our square root, paying careful attention to the negative values, we have 𝑡=−3±√9+800−10=−3±√809−10=3±√80910.

Therefore, our two solutions are 𝑡=3+√80910𝑡=3−√80910.or

Using a calculator, we can evaluate the 𝑡 values as 𝑡=−2.544𝑡=3.144.or

We can round to the nearest tenth of a second, giving us 𝑡=−2.5𝑡=3.1.or

As the time in seconds must be a positive value, we can reject the negative value of −2.5. Hence, our final answer is that the stone reaches the sea at 3.1 seconds, to the nearest tenth of a second.

How To: Deriving the Quadratic Formula

Suppose we have a quadratic equation in the general form 𝑎𝑥+𝑏𝑥+𝑐=0, where 𝑎, 𝑏, and 𝑐 are constants. We can start our process of solving this equation by dividing all the terms by 𝑎, which gives us 𝑥+𝑏𝑎𝑥+𝑐𝑎=0.

To begin isolating the 𝑥- and 𝑥-terms, we subtract 𝑐𝑎 from both sides of the equation, giving us 𝑥+𝑏𝑎𝑥=−𝑐𝑎.

On the left-hand side of the equation, we can now follow the process for completing the square. To complete the square of a quadratic, we need to write the quadratic in the form of a binomial squared.

Recall that we can rewrite an expression 𝑐+2𝑐𝑑, with constants 𝑐 and 𝑑, as 𝑐+2𝑐𝑑=(𝑐+𝑑)−𝑑.

So, comparing the left-hand side of our equation, we can equate 𝑐=𝑥 and 𝑑=𝑏2𝑎.

Therefore, we can substitute 𝑥+𝑏𝑎𝑥=𝑥+𝑏2𝑎−𝑏2𝑎 into our equation, giving us 𝑥+𝑏2𝑎−𝑏2𝑎=−𝑐𝑎.

Expanding 𝑏2𝑎, we have 𝑥+𝑏2𝑎−𝑏4𝑎=−𝑐𝑎.

Adding 𝑏4𝑎 to both sides gives us 𝑥+𝑏2𝑎=𝑏4𝑎−𝑐𝑎.

Using 4𝑎 as a common denominator on the right-hand side, we can simplify to give 𝑥+𝑏2𝑎=𝑏−4𝑎𝑐4𝑎.

We can now take the square root of both sides, remembering the positive and negative values, giving us 𝑥+𝑏2𝑎=±𝑏−4𝑎𝑐4𝑎𝑥+𝑏2𝑎=±√𝑏−4𝑎𝑐2𝑎.

Next, we subtract 𝑏2𝑎 from both sides, which gives us 𝑥=−𝑏2𝑎±√𝑏−4𝑎𝑐2𝑎.

As both terms on the right-hand side have the same denominator, we can write this in the simplified form as 𝑥=−𝑏±√𝑏−4𝑎𝑐2𝑎.

And so we have derived the quadratic formula, which gives a concise solution for all quadratic equations in the form 𝑎𝑥+𝑏𝑥+𝑐=0.

Key Points

  • If we have a quadratic equation in the form 𝑎𝑥+𝑏𝑥+𝑐=0, where 𝑎, 𝑏, and 𝑐 are constants and 𝑎≠0, we can use the quadratic formula, 𝑥=−𝑏±√𝑏−4𝑎𝑐2𝑎, to solve for 𝑥. We use the quadratic formula to solve any quadratic equation, even if the equation is not in 𝑥.
  • When we solve a quadratic equation, 𝑎𝑥+𝑏𝑥+𝑐=0, to find the values of 𝑥, we are finding the values of 𝑥 for which 𝑦=0. If we were to draw a graph of our equation, the 𝑥-value found by solving using the quadratic formula would give us the points where the graph crosses the 𝑥-axis.
  • When using the quadratic formula, there may be occasions where the value given by 𝑏−4𝑎𝑐, the discriminant of the equation, is less than zero. That is, 𝑏−4𝑎𝑐<0. This would mean that in the formula we would be attempting to take the square root of a negative value, so the solution would not be a real number. At this point, the quadratic formula would cease to work. If we think in terms of a graph of this equation, a discriminant where 𝑏−4𝑎𝑐<0 means that the graph would not pass through the 𝑥-axis. So, in these cases, there are no real values of 𝑥 for which 𝑦=0.

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