Explainer: Solving Quadratic Equations: Quadratic Formula

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 π‘Žπ‘₯+𝑏π‘₯+𝑐=0, where π‘Ž, 𝑏, and 𝑐 are constants and π‘Žβ‰ 0, then π‘₯=βˆ’π‘Β±βˆšπ‘βˆ’4π‘Žπ‘2π‘Ž.

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 π‘Ž, 𝑏, and 𝑐 from our equation into the quadratic formula. In the box below, we detail how to use the quadratic formula.

How to Use the Quadratic Formula

  • We first need to have an equation in the form π‘Žπ‘₯+𝑏π‘₯+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 π‘Žπ‘₯+𝑏π‘₯+𝑐=0, with the π‘₯, π‘₯, and constant terms in that order. Writing it in this form can help us ensure that we take the correct π‘Ž, 𝑏, and 𝑐 values. It is also useful to list our values π‘Ž=, 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 π‘Ž, 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 βˆ’π‘₯+7π‘₯+1=0.

Answer

An equation of the form π‘Žπ‘₯+𝑏π‘₯+𝑐=0 can be solved using the quadratic formula π‘₯=βˆ’π‘Β±βˆšπ‘βˆ’4π‘Žπ‘2π‘Ž.

In the equation βˆ’π‘₯+7π‘₯+1=0, we have π‘Ž=βˆ’1, 𝑏=7, and 𝑐=1. Therefore, we can substitute these values into the formula and simplify the terms, giving us π‘₯=βˆ’7±√7βˆ’4(βˆ’1)(1)2(βˆ’1).

Simplifying the terms, paying careful attention to the negative signs, we have π‘₯=βˆ’7±√49+4βˆ’2=βˆ’7±√53βˆ’2=7±√532.

Hence, we have the solution set 7βˆ’βˆš532,7+√532

Example 2: Solving an Equation Using the Quadratic Formula

Find the solution set of the equation 3π‘₯+4(π‘₯+1)=0, giving values in ℝ to one decimal place.

Answer

In order to begin solving this, we first expand the brackets to give us an equation in the form π‘Žπ‘₯+𝑏π‘₯+𝑐=0: 3π‘₯+4(π‘₯+1)=03π‘₯+4π‘₯+4=0.

This equation can now be solved using the quadratic formula π‘₯=βˆ’π‘Β±βˆšπ‘βˆ’4π‘Žπ‘2π‘Ž.

In the equation 3π‘₯+4π‘₯+4=0, we have the values π‘Ž=3, 𝑏=4, and 𝑐=4. Therefore, we can substitute these values into the formula, giving us π‘₯=βˆ’4±√4βˆ’(4)(3)(4)2(3)=βˆ’4±√16βˆ’486=βˆ’4Β±βˆšβˆ’326.

However, we can see that our calculation includes taking the square root of a negative number, βˆšβˆ’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 3π‘₯+4(π‘₯+1)=0. Hence, the solution set of the equation is the empty set βˆ….

It is important to note that although the quadratic formula breaks down when the discriminant, π‘βˆ’4π‘Žπ‘οŠ¨, of the calculation is less than zero, it simply means that the graph of this function would not pass through the π‘₯-axis, meaning that there are no solutions for π‘₯. If we were to draw a graph of 𝑦=3π‘₯+4(π‘₯+1), it would look as below.

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

Given that π‘₯=βˆ’1 is one of the roots of the equation βˆ’3π‘₯βˆ’9π‘₯+π‘˜=0, find the other root and the value of π‘˜.

Answer

We find the roots of an equation by calculating what the values of π‘₯ are when the 𝑦-value is equal to zero. When we have an equation in the form π‘Žπ‘₯+𝑏π‘₯+𝑐=0, where π‘Ž, 𝑏, and 𝑐 are constants, we can find the the values of π‘₯, the roots, by using the quadratic formula π‘₯=βˆ’π‘Β±βˆšπ‘βˆ’4π‘Žπ‘2π‘Ž.

Although we do not yet know the value of π‘˜ in the equation βˆ’3π‘₯βˆ’9π‘₯+π‘˜=0, we can substitute in the values π‘Ž=βˆ’3, 𝑏=βˆ’9, and 𝑐=π‘˜ into the quadratic formula, giving us π‘₯=9±√(βˆ’9)βˆ’(4)(βˆ’3)(π‘˜)2(βˆ’3).

Simplifying our terms, we have π‘₯=9±√81+12π‘˜βˆ’6.

We were given that one of the roots was βˆ’1, so we can substitute π‘₯=βˆ’1 into our equation, giving βˆ’1=9±√81+12π‘˜βˆ’6.

Multiplying both sides of the equation by βˆ’6, we have 6=9±√81+12π‘˜.

Subtracting 9 from both sides gives us βˆ’3=±√81+12π‘˜.

Squaring both sides, we have 9=81+12π‘˜.

We can then subtract 81 from both sides and simplify the terms in order to find the value of π‘˜, which gives us 9βˆ’81=12π‘˜βˆ’72=12π‘˜π‘˜=βˆ’6.

We can substitute π‘˜=βˆ’6 into our earlier workings where π‘₯=9±√81+12π‘˜βˆ’6. This gives us π‘₯=9±√81βˆ’72βˆ’6=9±√9βˆ’6=9Β±3βˆ’6.

So our solutions are π‘₯=βˆ’126π‘₯=βˆ’66π‘₯=βˆ’2π‘₯=βˆ’1.oror

Therefore, the roots of the equation are π‘₯=βˆ’2 and π‘₯=βˆ’1. As we were given the root π‘₯=βˆ’1, this means that our answer is π‘₯=βˆ’2 and π‘˜=βˆ’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 π‘₯, 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=5Γ—12=60.

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

We can expand the brackets and rearrange the equation into the form π‘Žπ‘₯+𝑏π‘₯+𝑐=0, with constants π‘Ž, 𝑏, and 𝑐, giving us 60+17π‘₯+π‘₯=120 which we can rewrite as π‘₯+17π‘₯βˆ’60=0.

Quadratic equations in the form π‘Žπ‘₯+𝑏π‘₯+𝑐=0 can be solved using the quadratic formula π‘₯=βˆ’π‘Β±βˆšπ‘βˆ’4π‘Žπ‘2π‘Ž.

We can substitute the values π‘Ž=1, 𝑏=17, and 𝑐=βˆ’60 into the quadratic formula and simplify, giving us π‘₯=βˆ’17±√17βˆ’4(1)(βˆ’60)2Γ—1.

Simplifying the terms, paying careful attention to the negative signs within the square root, we have π‘₯=βˆ’17±√289+2402=βˆ’17±√5292=βˆ’17Β±232.

Therefore, π‘₯=βˆ’17+232π‘₯=βˆ’17βˆ’232π‘₯=62π‘₯=βˆ’402.oror

So, π‘₯=3π‘₯=βˆ’20.or

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

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

Therefore, π‘₯=βˆ’20π‘₯=3.or

In the same way as above, we require a positive dimension and reject the negative value of π‘₯. So the required answer is π‘₯=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 π‘₯. We follow the same process of using the quadratic formula with the coefficients for values π‘Ž, 𝑏, and 𝑐 in the same way as we do for a quadratic in π‘₯.

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, β„Ž, above sea level at any time, 𝑑 seconds, is given by the formula β„Ž=3π‘‘βˆ’5𝑑+40.

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

Answer

The equation β„Ž=3π‘‘βˆ’5𝑑+40 is a quadratic equation in 𝑑. We can rewrite it into the form β„Ž=βˆ’5𝑑+3𝑑+40.

When the stone reaches the sea, this will mean that the height above sea level is zero, β„Ž=0, so this gives us βˆ’5𝑑+3𝑑+40=0.

Recall that a quadratic equation in π‘₯, in the form π‘Žπ‘₯+𝑏π‘₯+𝑐=0, can be solved using the quadratic formula π‘₯=βˆ’π‘Β±βˆšπ‘βˆ’4π‘Žπ‘2π‘Ž.

Since we have the form π‘Žπ‘‘+𝑏𝑑+𝑐=0, we can solve with the quadratic formula written as 𝑑=βˆ’π‘Β±βˆšπ‘βˆ’4π‘Žπ‘2π‘Ž.

Now we can substitute the valuesπ‘Ž=βˆ’5, 𝑏=3, and 𝑐=40 into the quadratic formula and simplify, which gives us 𝑑=βˆ’3±√3βˆ’4(βˆ’5)(40)2(βˆ’5).

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

Therefore, our two solutions are 𝑑=3+√80910𝑑=3βˆ’βˆš80910.or

Using a calculator, we can evaluate the 𝑑 values as 𝑑=βˆ’2.544𝑑=3.144.or

We can round to the nearest tenth of a second, giving us 𝑑=βˆ’2.5𝑑=3.1.or

As the time in seconds must be a positive value, we can reject the negative value of βˆ’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 Derive the Quadratic Formula

Suppose we have a quadratic equation in the general form π‘Žπ‘₯+𝑏π‘₯+𝑐=0, where π‘Ž, 𝑏, and 𝑐 are constants. We can start our process of solving this equation by dividing all the terms by π‘Ž, which gives us π‘₯+π‘π‘Žπ‘₯+π‘π‘Ž=0.

To begin isolating the π‘₯- and π‘₯-terms, we subtract π‘π‘Ž from both sides of the equation, giving us π‘₯+π‘π‘Žπ‘₯=βˆ’π‘π‘Ž.

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+2π‘π‘‘οŠ¨, with constants 𝑐 and 𝑑, as 𝑐+2𝑐𝑑=(𝑐+𝑑)βˆ’π‘‘.

So, comparing the left-hand side of our equation, we can equate 𝑐=π‘₯ and 𝑑=𝑏2π‘Ž.

Therefore, we can substitute π‘₯+π‘π‘Žπ‘₯=ο€½π‘₯+𝑏2π‘Žο‰βˆ’ο€½π‘2π‘Žο‰οŠ¨οŠ¨οŠ¨ into our equation, giving us ο€½π‘₯+𝑏2π‘Žο‰βˆ’ο€½π‘2π‘Žο‰=βˆ’π‘π‘Ž.

Expanding 𝑏2π‘Žο‰οŠ¨, we have ο€½π‘₯+𝑏2π‘Žο‰βˆ’π‘4π‘Ž=βˆ’π‘π‘Ž.

Adding 𝑏4π‘ŽοŠ¨οŠ¨ to both sides gives us ο€½π‘₯+𝑏2π‘Žο‰=𝑏4π‘Žβˆ’π‘π‘Ž.

Using 4π‘ŽοŠ¨ as a common denominator on the right-hand side, we can simplify to give ο€½π‘₯+𝑏2π‘Žο‰=π‘βˆ’4π‘Žπ‘4π‘Ž.

We can now take the square root of both sides, remembering the positive and negative values, giving us π‘₯+𝑏2π‘Ž=Β±ο„žπ‘βˆ’4π‘Žπ‘4π‘Žπ‘₯+𝑏2π‘Ž=Β±βˆšπ‘βˆ’4π‘Žπ‘2π‘Ž.

Next, we subtract 𝑏2π‘Ž from both sides, which gives us π‘₯=βˆ’π‘2π‘ŽΒ±βˆšπ‘βˆ’4π‘Žπ‘2π‘Ž.

As both terms on the right-hand side have the same denominator, we can write this in the simplified form as π‘₯=βˆ’π‘Β±βˆšπ‘βˆ’4π‘Žπ‘2π‘Ž.

And so we have derived the quadratic formula, which gives a concise solution for all quadratic equations in the form π‘Žπ‘₯+𝑏π‘₯+𝑐=0.

Key Points

  • If we have a quadratic equation in the form π‘Žπ‘₯+𝑏π‘₯+𝑐=0, where π‘Ž, 𝑏, and 𝑐 are constants and π‘Žβ‰ 0, we can use the quadratic formula, π‘₯=βˆ’π‘Β±βˆšπ‘βˆ’4π‘Žπ‘2π‘ŽοŠ¨, to solve for π‘₯. We use the quadratic formula to solve any quadratic equation, even if the equation is not in π‘₯.
  • When we solve a quadratic equation, π‘Žπ‘₯+𝑏π‘₯+𝑐=0, to find the values of π‘₯, we are finding the values of π‘₯ for which 𝑦=0. If we were to draw a graph of our equation, the π‘₯-value found by solving using the quadratic formula would give us the points where the graph crosses the π‘₯-axis.
  • When using the quadratic formula, there may be occasions where the value given by π‘βˆ’4π‘Žπ‘οŠ¨, the discriminant of the equation, is less than zero. That is, π‘βˆ’4π‘Žπ‘<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 π‘βˆ’4π‘Žπ‘<0 means that the graph would not pass through the π‘₯-axis. So, in these cases, there are no real values of π‘₯ for which 𝑦=0.

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