Lesson Explainer: Solving Quadratic Equations: Quadratic Formula | Nagwa Lesson Explainer: Solving Quadratic Equations: Quadratic Formula | Nagwa

Lesson Explainer: Solving Quadratic Equations: Quadratic Formula Mathematics

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

Imagine we are asked to solve the quadratic equation 𝑥+5𝑥+3=0; there are several different ways we could go about this. For example, we could try to factor the quadratic by looking at the factor pairs of 3 and trying to make them sum to 5. However, 3 only has one pair of factors, 1 and 3, which do not sum to 5. Hence, we can immediately see this is not possible, at least for integer values, so instead, we can do this by completing the square.

We want to write the quadratic expression in the form (𝑥+𝑏′)+𝑐′. We do this by halving the coefficient of 𝑥 to get 𝑏′, so 𝑏′=52. Then, 𝑐′ is 3−(𝑏′), so 𝑐′=3−52=−134.

Hence, 𝑥+5𝑥+3=𝑥+52−134.

Substituting this expression into the equation we want to solve yields 𝑥+52−134=0.

We can now solve this equation for 𝑥; we first add 134 to both sides to get 𝑥+52=134.

We then take the square root of both sides of the equation, where we will get a positive and a negative root: 𝑥+52=±134𝑥+52=±√132.

Finally, we subtract 52 from both sides of the equation, giving us the two roots: 𝑥=−52±√132.

We can apply this same process to find the roots of any quadratic; let’s say we want to solve 𝑎𝑥+𝑏𝑥+𝑐=0, where 𝑎≠0. We first divide the equation through by 𝑎 to get 𝑥+𝑏𝑎𝑥+𝑐𝑎=0.

Next, we complete the square on the left-hand side of the equation to get 𝑥+𝑏𝑎𝑥+𝑐𝑎=𝑥+𝑏2𝑎+𝑐𝑎−𝑏2𝑎.

Hence, after distributing the exponent and rearranging, our equation becomes 𝑥+𝑏2𝑎−𝑏4𝑎+𝑐𝑎=0.

We then rearrange to get 𝑥+𝑏2𝑎=𝑏4𝑎−𝑐𝑎.

We add the terms on the right-hand side of the equation by rewriting the terms with a common denominator: 𝑥+𝑏2𝑎=𝑏4𝑎−𝑐𝑎×4𝑎4𝑎𝑥+𝑏2𝑎=𝑏−4𝑎𝑐4𝑎.

We then want to take the square root of both sides of the equation. We note that this step is only valid if the right-hand side of the equation is nonnegative. We find a positive and a negative root, this gives us 𝑥+𝑏2𝑎=±𝑏−4𝑎𝑐4𝑎.

We distribute the root over the numerator and denominator and then rearrange the equation for 𝑥, giving us 𝑥+𝑏2𝑎=±√𝑏−4𝑎𝑐√4𝑎𝑥=−𝑏2𝑎±√𝑏−4𝑎𝑐2𝑎𝑥=−𝑏±√𝑏−4𝑎𝑐2𝑎.

Therefore, if these values exist, then they are the roots of the quadratic equation in terms of the coefficients 𝑎, 𝑏, and 𝑐. We can summarize this result as follows.

Formula: The Quadratic Formula

The roots of the quadratic equation 𝑎𝑥+𝑏𝑥+𝑐=0, are given by 𝑥=−𝑏±√𝑏−4𝑎𝑐2𝑎.

We can use this formula to solve quadratic equations without factoring or completing the square; all that is required is to substitute the values of the coefficients into the formula. This allows us to easily solve some quadratic equations that were previously difficult to solve because the quadratic was not easily factorable.

This gives us the following process to attempt to solve quadratic equations.

How To: Solving Quadratic Equations Using the Quadratic Formula

  1. We first need to ensure that the equation is in the form 𝑎𝑥+𝑏𝑥+𝑐=0, where 𝑎 is nonzero. We may need to distribute parentheses or rearrange 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=, and 𝑐=, so that they can be referenced when substituting.
  2. Next, we should write down the quadratic formula in full. This will help avoid errors and ensure that the positive and negative values are consistently applied throughout our working.
  3. We then substitute the values of 𝑎, 𝑏, and 𝑐 into the quadratic formula. It is important to write any negative number substitutions in parentheses as this will help maintain the accuracy of the positive and negative values as we go through the steps of the solution.
  4. Finally, we will get two values (one for each of the ± operations), and both of these values are solutions to the equation.

Let’s see some examples of how to apply the quadratic formula to solve some equations.

Example 1: Finding the Solutions to a Quadratic Equation

Find the solution set of the equation 𝑥−3𝑥+1=0, giving values correct to two decimal places.

Answer

We are asked to solve a quadratic equation. If we attempted to solve this equation by factoring, we would need two numbers that multiply to give 1 and add to give −3, and two such numbers are difficult to find. Another thing worth noting is that the question says to give the roots to two decimal places, this indicates that the roots may be difficult to find exactly. In both cases, we should conclude that the quadratic formula will be a good method to try.

We recall that the quadratic formula tells us that the roots of the quadratic equation 𝑎𝑥+𝑏𝑥+𝑐=0 are given by 𝑥=−𝑏±√𝑏−4𝑎𝑐2𝑎, provided 𝑎≠0.

The values of 𝑎, 𝑏, and 𝑐 are the coefficients of the quadratic expression, where 𝑎=1, 𝑏=−3, and 𝑐=1. Substituting these values into the formula and simplifying gives us 𝑥=−(−3)±(−3)−4(1)(1)2(1)=3±√9−42=3±√52.

This gives us an expression for each root (one for each of the + or – operations). We have 𝑥=3+√52 and 𝑥=3−√52 as the roots, and we can evaluate these to two decimal places to get 𝑥≈2.62 and 𝑥≈0.38.

Finally, we note that the solution set of an equation is the set of all solutions to the equation, so we want the set of all roots given to two decimal places, which is {2.62,0.38}.

Example 2: Solving an Equation Using the Quadratic Formula

Solve the equation −𝑥+7𝑥+1=0.

Answer

We are asked to solve a quadratic equation, which if we attempted to solve by factoring, we would need two numbers that multiply to give 1 and add to give 7, and two such numbers are difficult to find. Instead, we will apply the quadratic formula, which we recall tells us that the roots of the quadratic equation 𝑎𝑥+𝑏𝑥+𝑐=0 are given by 𝑥=−𝑏±√𝑏−4𝑎𝑐2𝑎.

We have two choices; we can directly apply the quadratic formula to this quadratic equation, or we can rearrange the equation into standard form first by multiplying through by −1. This would give us 𝑥−7𝑥−1=0.

Thus, 𝑎=1, 𝑏=−7, and 𝑐=−1. Substituting these values into the formula and simplifying gives us 𝑥=−(−7)±(−7)−4(1)(−1)2(1)=7±√49+42=7±√532.

Hence, the roots are 𝑥=7+√532 and 𝑥=7−√532. Therefore, the set of solutions to this equation is 7−√532,7+√532.

In our next few examples, we will see that we may need to rearrange the given equation into a quadratic equation in the correct form before applying the quadratic formula.

Example 3: Solving Quadratic Equations

Find the solution set of the equation 5𝑥(𝑥−6)−3(𝑥+4)+3=0, giving values to one decimal place.

Answer

We cannot solve this equation in its current form, so we will start by distributing the parentheses and simplifying: 5𝑥(𝑥−6)−3(𝑥+4)+3=05𝑥−30𝑥−3𝑥−12+3=0.

Collecting like terms then yields 5𝑥−33𝑥−9=0.

This is a quadratic equation; we cannot factor this equation easily, so we will use the quadratic formula. We recall that the quadratic formula tells us that the roots of the quadratic equation 𝑎𝑥+𝑏𝑥+𝑐=0 are given by 𝑥=−𝑏±√𝑏−4𝑎𝑐2𝑎.

In this quadratic, 𝑎=5, 𝑏=−33, and 𝑐=−9. Substituting these values into the formula and simplifying gives us 𝑥=−(−33)±(−33)−4(5)(−9)2(5)=33±√1089+18010=33±√126910.

Therefore, the roots of the quadratic are 𝑥=33−√126910 and 𝑥=33+√126910. We can evaluate these expressions to one decimal place to get 𝑥≈−0.3 and 𝑥≈6.9.

Finally, we note that the solution set of an equation is the set of all solutions to that equation, so we want the set of all roots given to one decimal place, which is {6.9,−0.3}.

Example 4: Rearranging an Equation to Use the Quadratic Formula

Find the solution set of the equation 18𝑥+5𝑥=1, giving values to three decimal places.

Answer

We cannot solve the equation in its current form, so we will need to rearrange the equation into a form that we can solve. To do this, we first want to multiply through by 𝑥 to remove the denominators in the equation.

Before we do this, we need to consider that multiplying through by a variable can introduce extra solutions. In particular, we may introduce the solution 𝑥=0, since both sides of the equation would have a factor of 𝑥. However, we already know that 𝑥≠0 since we are dividing by 𝑥 in the equation, so we can just ignore this solution if it occurs.

Multiplying the equation through by 𝑥, we have 𝑥⋅18𝑥+5𝑥=𝑥⋅(1)18⋅𝑥𝑥+5⋅𝑥𝑥=𝑥.

We then cancel the shared factors on the left-hand side of the equation to get 18+5𝑥=𝑥.

Finally, we rearrange this into the following quadratic equation: 𝑥−5𝑥−18=0.

We now have a quadratic equation that we cannot easily solve by factoring, so we will use the quadratic formula.

We recall that the quadratic formula tells us that the roots of the quadratic equation 𝑎𝑥+𝑏𝑥+𝑐=0 are given by 𝑥=−𝑏±√𝑏−4𝑎𝑐2𝑎.

In this quadratic, we have 𝑎=1, 𝑏=−5, and 𝑐=−18. Substituting these values into the formula and simplifying, we have 𝑥=−(−5)±(−5)−4(1)(−18)2(1)=5±√25+722=5±√972.

This gives us two roots: 𝑥=5+√972 and 𝑥=5−√972. To three decimal places, these roots are 𝑥≈7.424 and 𝑥≈−2.424.

Finally, we note that the solution set of an equation is the set of all solutions to the equation, so we want the set of all roots given to three decimal places, which is {7.424,−2.424}.

In our final two examples, we will use given properties of quadratic equations and the quadratic formula to determine the value of an unknown.

Example 5: Finding an Unknown in a Quadratic Equation Using a Root

Given that 𝑥=−2 is a root of the equation 𝑥−4𝑚𝑥−𝑚−6=0, find the set of possible values of 𝑚.

Answer

We first note that we are given a quadratic equation in 𝑥. To determine the value of 𝑚, we might be tempted to directly apply the quadratic formula to this equation; however, there is an easier method where we first use the fact that 𝑥=−2 is a root.

Since 𝑥=−2 is a root, it satisfies the equation; hence, we can substitute this value into the equation to get (−2)−4𝑚(−2)−𝑚−6=0.

Distributing and simplifying, we get 4+8𝑚−𝑚+6=0−𝑚+8𝑚+10=0𝑚−8𝑚−10=0.

This is then a quadratic equation in 𝑚, and so we can solve for 𝑚 by applying the quadratic formula. The equation 𝑎𝑚+𝑏𝑚+𝑐=0 has solutions given by 𝑚=−𝑏±√𝑏−4𝑎𝑐2𝑎.

In this quadratic, we have 𝑎=1, 𝑏=−8, and 𝑐=−10. Substituting these values in and simplifying, we have 𝑚=−(−8)±(−8)−4(1)(−10)2(1)=8±√64+402=8±√1042.

We can simplify further by simplifying the radical: 𝑚=8±√4×262=8±2√262=4±√26.

This gives two possible values of 𝑚, either 𝑚=4+√26 or 𝑚=4−√26. Hence, the set of possible values of 𝑚 is {4−√26,4+√26}.

In our final example, we will once again use a root of the equation to find the value of an unknown in a quadratic equation and the full solution set.

Example 6: Finding an Unknown in a Quadratic Equation Using the Relation between Its Coefficients and Roots

The sum of the roots of the equation 4𝑥+𝑘𝑥−4=0 is −1. Find the value of 𝑘 and the solution set of the equation.

Answer

There are a few ways we can determine the value of 𝑘; we will use the quadratic formula. First, since we are told that the equation has roots, we know that we can apply the quadratic formula, which tells us that the roots of the equation 𝑎𝑥+𝑏𝑥+𝑐=0 will be 𝑥=−𝑏±√𝑏−4𝑎𝑐2𝑎. In this quadratic equation, we have 𝑎=4, 𝑏=𝑘, and 𝑐=−4. Substituting these values in and simplifying, we have 𝑥=−𝑘±√𝑘−4(4)(−4)2(4)=−𝑘±√𝑘+648.

So, we have the roots 𝑥=−𝑘+√𝑘+648 and 𝑥=−𝑘−√𝑘+648. The sum of these roots is then given by −𝑘+√𝑘+648+−𝑘−√𝑘+648=−𝑘+√𝑘+64−𝑘−√𝑘+648=−2𝑘8=−𝑘4.

Since the sum of the roots is −1, we must have −1=−𝑘4𝑘=4.

We could substitute 𝑘=4 back into the original equation; however, we already found expressions for the roots: 𝑥=−𝑘±√𝑘+648. Substituting 𝑘=4 and simplifying yields 𝑥=−4±√4+648=−4±√808=−4±√16×58=−4±4√58=−1±√52.

Hence, the roots are 𝑥=−1−√52 and 𝑥=−1+√52.

Therefore, the value of 𝑘 is 4, and the solution set of the equation is −1−√52,−1+√52.

In the previous example, we showed that the sum of the roots of a given quadratic was equal to the negative of the quotients of the coefficients of 𝑥 and 𝑥. It is worth noting that a similar result is true in general; since the roots are 𝑥=−𝑏±√𝑏−4𝑎𝑐2𝑎, the sum of the roots is given by −𝑏+√𝑏−4𝑎𝑐2𝑎+−𝑏−√𝑏−4𝑎𝑐2𝑎=−2𝑏2𝑎=−𝑏𝑎.

Hence, the sum of the roots is always equal to the negative of the quotients of the coefficients of 𝑥 and 𝑥.

Let’s finish by recapping some of the important points of this explainer.

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.
  • The quadratic formula can be used for quadratics that are too difficult to factor.
  • We can apply the quadratic formula to quadratics in any variable.

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