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

Lesson Explainer: Solving Quadratic Equations: Quadratic Formula Mathematics • First Year of Secondary School

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

Recall that a quadratic equation is an equation with one variable, where the highest order of any term is 2. This is made more explicit in the definition below.

Definition: Quadratic Equation

The equation π‘Žπ‘₯+𝑏π‘₯+𝑐=0, π‘Žβ‰ 0, with variable π‘₯ and constants π‘Ž, 𝑏, and 𝑐, is called a quadratic equation (or second-degree equation).

To solve quadratic equations, we can use the following methods:

  • Factoring
  • Completing the square
  • Using the quadratic formula
  • Solving graphically

So far, we have met factoring and completing the square. We will use completing the square to derive the quadratic formula, which is the method we will be focusing on in this explainer.

If we have some quadratic equation of the form π‘Žπ‘₯+𝑏π‘₯+𝑐=0, π‘Žβ‰ 0, with variable π‘₯ and constants π‘Ž, 𝑏, and 𝑐, then using completing the square, we can rearrange and solve for π‘₯ as follows.

First, we will divide through by π‘Ž, the coefficient of π‘₯: π‘Žπ‘₯+𝑏π‘₯+𝑐=0π‘₯+π‘π‘Žπ‘₯+π‘π‘Ž=0.

Next, we will subtract π‘π‘Ž from both sides: π‘₯+π‘π‘Žπ‘₯=βˆ’π‘π‘Ž.

Then, to make a perfect square, we will add 𝑏2π‘Žο‰οŠ¨ to both sides: π‘₯+π‘π‘Žπ‘₯+𝑏2π‘Žο‰=βˆ’π‘π‘Ž+𝑏2π‘Žο‰.

Since the left-hand side of the equation is in the form π‘₯+2𝑑π‘₯+π‘‘οŠ¨οŠ¨, a perfect square, then we can write this as (π‘₯+𝑑): ο€½π‘₯+𝑏2π‘Žο‰=βˆ’π‘π‘Ž+𝑏2π‘Žο‰.

Next, we will rearrange to get π‘₯ as the subject. First, we square root both sides (taking both the positive and negative square roots): ο€½π‘₯+𝑏2π‘Žο‰=βˆ’π‘π‘Ž+𝑏2π‘Žο‰π‘₯+𝑏2π‘Ž=Β±ο„Ÿβˆ’π‘π‘Ž+𝑏2π‘Žο‰.

Next, we isolate π‘₯: π‘₯=βˆ’π‘2π‘ŽΒ±ο„Ÿβˆ’π‘π‘Ž+𝑏2π‘Žο‰.

We then simplify further by expanding the brackets in the radical and rearranging slightly: π‘₯=βˆ’π‘2π‘ŽΒ±ο„Ÿο€½π‘2π‘Žο‰βˆ’π‘π‘Ž=βˆ’π‘2π‘ŽΒ±ο„žπ‘4π‘Žβˆ’π‘π‘Ž.

We can now write the whole expression in the radical over a common denominator: π‘₯=βˆ’π‘2π‘ŽΒ±ο„žπ‘4π‘Žβˆ’4π‘Žπ‘4π‘Ž=βˆ’π‘2π‘ŽΒ±ο„žπ‘βˆ’4π‘Žπ‘4π‘Ž.

Since the denominator in the radical is a perfect square, then we can square root this and write it outside of the radical: π‘₯=βˆ’π‘2π‘ŽΒ±βˆšπ‘βˆ’4π‘Žπ‘2π‘Ž.

Combining the two terms, since they have the same denominator, we then get π‘₯=βˆ’π‘Β±βˆšπ‘βˆ’4π‘Žπ‘2π‘Ž.

So in its final form, π‘₯=βˆ’π‘Β±βˆšπ‘βˆ’4π‘Žπ‘2π‘ŽοŠ¨ is the quadratic formula, which is used for finding solutions of quadratic equations of the form π‘Žπ‘₯+𝑏π‘₯+𝑐=0, π‘Žβ‰ 0. This is stated in the definition below.

Definition: The Quadratic Formula

To solve a quadratic equation in the form π‘Žπ‘₯+𝑏π‘₯+𝑐=0, π‘Žβ‰ 0, with variable π‘₯ and constants π‘Ž, 𝑏, and 𝑐 we can use the quadratic formula to solve for π‘₯: π‘₯=βˆ’π‘Β±βˆšπ‘βˆ’4π‘Žπ‘2π‘Ž.

Whenever we want to use the quadratic formula, we need to ensure that our quadratic equation is equal to zero, with it in expanded form and simplified as much as possible, so that it is in the form π‘Žπ‘₯+𝑏π‘₯+𝑐=0, π‘Žβ‰ 0. Then, we need to identify what π‘Ž, 𝑏, and 𝑐 each are. Following that, we can substitute into the quadratic formula and solve for π‘₯. There are usually two values of π‘₯, corresponding to the positive and negative square roots of π‘βˆ’4π‘Žπ‘οŠ¨, but sometimes there is only one, or even no real solutions, depending on the values of π‘Ž, 𝑏, and 𝑐.

We will discuss how to solve a quadratic equation which is already in the form π‘Žπ‘₯+𝑏π‘₯+𝑐=0, π‘Žβ‰ 0 in our first example.

Example 1: Solving Quadratic Equations Using the Quadratic Formula

Find the solution set of the equation 6π‘₯βˆ’8π‘₯+1=0, giving values to two decimal places.

Answer

As the equation 6π‘₯βˆ’8π‘₯+1=0 is a quadratic equation, then we can use one of the methods for solving quadratics. In this case, we are going to use the quadratic formula in order to solve. Recall, that for a quadratic equation in the form π‘Žπ‘₯+𝑏π‘₯+𝑐=0, π‘Žβ‰ 0, then π‘₯=βˆ’π‘Β±βˆšπ‘βˆ’4π‘Žπ‘2π‘Ž.

Since 6π‘₯βˆ’8π‘₯+1=0 is already in the same form as π‘Žπ‘₯+𝑏π‘₯+𝑐=0 since it is equal to zero, fully simplified, and written in descending powers of π‘₯, then we can identify the values of π‘Ž, 𝑏, and 𝑐: π‘Ž=6,𝑏=βˆ’8,𝑐=1.and

Substituting these values into the quadratic formula and solving for π‘₯, we get π‘₯=βˆ’π‘Β±βˆšπ‘βˆ’4π‘Žπ‘2π‘Ž=βˆ’(βˆ’8)±(βˆ’8)βˆ’4(6)(1)2(6)=8±√64βˆ’2412=8±√4012=8Β±2√1012=4±√106.

As we are asked to give the solution set to 2 decimal places, then evaluating π‘₯, we get π‘₯=1.19…0.14….or

So, the solution set of the equation 6π‘₯βˆ’8π‘₯+1=0 is {0.14,1.19} correct to 2 decimal places.

In the next example, we will consider how to solve a quadratic equation that is not in the form π‘Žπ‘₯+𝑏π‘₯+𝑐=0 initially but, by rearranging, can be put in this form and then solved using the quadratic formula.

Example 2: Solving Quadratic Equations Using the Quadratic Formula

Find the solution set of π‘₯βˆ’6(π‘₯βˆ’1)=2 in ℝ, giving values to two decimal places.

Answer

As the equation π‘₯βˆ’6(π‘₯βˆ’1)=2 contains a term with π‘₯, then this is likely to be a quadratic equation. To check, we can simplify by expanding brackets and making the equation equal to zero, as follows: π‘₯βˆ’6(π‘₯βˆ’1)=2π‘₯βˆ’6π‘₯+6=2π‘₯βˆ’6π‘₯+4=0.

As the highest power in the equation is 2, then we can see this is a quadratic equation. Since it is now written in the form π‘Žπ‘₯+𝑏π‘₯+𝑐=0, π‘Žβ‰ 0 then we can apply the quadratic formula, which states π‘₯=βˆ’π‘Β±βˆšπ‘βˆ’4π‘Žπ‘2π‘Ž.

We can see that for π‘₯βˆ’6π‘₯+4=0, π‘Ž=1, 𝑏=βˆ’6, and 𝑐=4. Substituting this into the quadratic formula, we get π‘₯=βˆ’(βˆ’6)±(βˆ’6)βˆ’4(1)(4)2(1).

Simplifying, we get π‘₯=6±√36βˆ’162=6±√202=6Β±2√52=3±√5.

As we are asked to give the solution set to 2 decimal places, then evaluating π‘₯, we get π‘₯=5.24…π‘₯=0.76….or

So, the solution set of π‘₯βˆ’6(π‘₯βˆ’1)=2 correct to 2 decimal places is {0.76,5.24}.

We can use the solutions of quadratic equations to find unknown parts of the equation, such as coefficients or constants. We can do this by substituting known and unknown parts into the quadratic formula and solving for the unknown part. We will explore how to do this in the next example.

Example 3: Finding Unknowns in Quadratic Equations Using the Quadratic Formula

Given that π‘₯=βˆ’2 is a root of the equation π‘₯βˆ’4π‘šπ‘₯βˆ’ο€Ήπ‘šβˆ’6=0, find the set of possible values of π‘š.

Answer

Since π‘₯βˆ’4π‘šπ‘₯βˆ’ο€Ήπ‘šβˆ’6=0 is of the form of a quadratic equation, then we can use the quadratic formula to find the values of π‘š.

The quadratic formula states that for an equation in the form π‘Žπ‘₯+𝑏π‘₯+𝑐=0, π‘Žβ‰ 0 then π‘₯=βˆ’π‘Β±βˆšπ‘βˆ’4π‘Žπ‘2π‘Ž.

We can see by comparing π‘₯βˆ’4π‘šπ‘₯βˆ’ο€Ήπ‘šβˆ’6=0 with π‘Žπ‘₯+𝑏π‘₯+𝑐=0 that π‘Ž=1, 𝑏=βˆ’4π‘š, and 𝑐=βˆ’ο€Ήπ‘šβˆ’6ο…οŠ¨. Since we also know that one of the roots of the equation is π‘₯=βˆ’2 then we can substitute π‘Ž, 𝑏, 𝑐 and π‘₯ into the quadratic formula: π‘₯=βˆ’π‘Β±βˆšπ‘βˆ’4π‘Žπ‘2π‘Žβˆ’2=βˆ’(βˆ’4π‘š)±(βˆ’4π‘š)βˆ’4(βˆ’(π‘šβˆ’6)(1))2(1).

Simplifying, we get βˆ’2=4π‘šΒ±βˆš16π‘š+4(π‘šβˆ’6)2=4π‘šΒ±βˆš16π‘š+4π‘šβˆ’242=4π‘šΒ±βˆš20π‘šβˆ’242.

We can simplify further by factoring out 4 and putting it outside the radical: βˆ’2=4π‘šΒ±βˆš4(5π‘šβˆ’6)2=4π‘šΒ±2√(5π‘šβˆ’6)2=2π‘šΒ±βˆš(5π‘šβˆ’6).

Next, we need to solve for π‘š. Since part of the equation contains π‘šοŠ¨, then once rearranged, it is likely to be the form of a quadratic equation. As such, we want to rearrange so it’s in the form π‘Žπ‘š+π‘π‘š+𝑐=0, π‘Žβ‰ 0, so that we can apply the quadratic formula again (but this time with different values for π‘Ž, 𝑏, and 𝑐).

Rearranging, we get βˆ’2=2π‘šΒ±βˆš(5π‘šβˆ’6)βˆ’2βˆ’2π‘š=±√(5π‘šβˆ’6)(βˆ’2βˆ’2π‘š)=5π‘šβˆ’64+8π‘š+4π‘š=5π‘šβˆ’60=5π‘šβˆ’4π‘šβˆ’8π‘šβˆ’4βˆ’6π‘šβˆ’8π‘šβˆ’10=0.

Now that the equation is in the form π‘Žπ‘š+π‘π‘š+𝑐=0, π‘Žβ‰ 0, we can find π‘Ž, 𝑏, and 𝑐. By comparing, we can see that π‘Ž=1, 𝑏=βˆ’8, and 𝑐=βˆ’10. Substituting into the quadratic formula, we get π‘š=βˆ’π‘Β±βˆšπ‘βˆ’4π‘Žπ‘2π‘Ž=βˆ’(βˆ’8)±(βˆ’8)βˆ’4(1)(βˆ’10)2(1).

Simplifying, we get π‘š=8±√64+402=8±√1042=8Β±2√262=4±√26.

Therefore, the possible values of π‘š are {4βˆ’βˆš26,4+√26}.

In addition to quadratic equations that contain quadratic terms, we can have equations that may not appear to be quadratic on first inspection but with some rearranging become quadratics. For example, 1π‘₯=4π‘₯ can be rearranged to give 1=4π‘₯, which is a quadratic equation. Therefore, we can solve equations, that once rearranged, become quadratics, and can do so using the quadratic formula. In our next example we will explore how to do this.

Example 4: Rearranging Equation to Solve Using the Quadratic Formula

Find the solution set of the equation βˆ’5βˆ’5π‘₯=1π‘₯ in ℝ, giving values to one decimal place.

Answer

In order to solve βˆ’5βˆ’5π‘₯=1π‘₯, it is helpful to remove any variables in the denominators first. To do this we need to multiply through by the highest power of π‘₯ in the denominator, which is π‘₯. Doing so gives us βˆ’5βˆ’5π‘₯=1π‘₯βˆ’5π‘₯βˆ’5π‘₯π‘₯=π‘₯π‘₯βˆ’5π‘₯βˆ’5π‘₯=1.

As the highest power of the equation is 2, then we can see this is a quadratic equation. To solve using the quadratic formula, we need to rearrange to make this in the form π‘Žπ‘₯+𝑏π‘₯+𝑐=0, π‘Žβ‰ 0. In this case, it is helpful to move the terms on the left-hand side to the right-hand side so that the coefficients are positive (but it does not necessarily matter) so that it is easier to do calculations later: βˆ’5π‘₯βˆ’5π‘₯=10=5π‘₯+5π‘₯+15π‘₯+5π‘₯+1=0.

Now that this is in the form π‘Žπ‘₯+𝑏π‘₯+𝑐=0, π‘Žβ‰ 0 we can apply the quadratic formula, which states π‘₯=βˆ’π‘Β±βˆšπ‘βˆ’4π‘Žπ‘2π‘Ž.

Comparing 5π‘₯+5π‘₯+1=0 with π‘Žπ‘₯+𝑏π‘₯+𝑐=0, we can see that π‘Ž=5, 𝑏=5, and 𝑐=1. Substituting into the quadratic formula, we get π‘₯=βˆ’5±√5βˆ’4(5)(1)2(5).

Simplifying, we get π‘₯=βˆ’5±√25βˆ’2010=βˆ’5±√510.

As the question requires us to find the solution set accurate to one decimal place, then we will evaluate this, giving us π‘₯=βˆ’0.28…π‘₯=βˆ’0.72.or

Therefore, the solution set for the equation βˆ’5βˆ’5π‘₯=1π‘₯ correct to one decimal place is {βˆ’0.3,βˆ’0.7}.

In addition to equations that can be rearranged to give a quadratic equation, we can have equations that are not quadratics themselves but can be solved using the quadratic formula as they are in a quadratic form. For example, π‘₯+2π‘₯+1=0οŠͺ is a quartic equation, as its highest power is 4, but as it takes the form of π‘Žπ‘¦+𝑏𝑦+𝑐=0, π‘Žβ‰ 0, where the variable 𝑦 represents π‘₯ or ο€Ήπ‘₯+2ο€Ήπ‘₯+1=0, then it can be solved using the quadratic formula since it has a quadratic form.

In the next example, we will consider how to solve a quartic equation by writing it in the form of a quadratic equation and using the quadratic formula.

Example 5: Using the Quadratic Formula to Solve a Quartic Equation

Using the quadratic formula, find all the solutions to π‘₯βˆ’10π‘₯+1=0οŠͺ.

Answer

To find the solutions of the equation π‘₯βˆ’10π‘₯+1=0οŠͺ using the quadratic formula, we need to write the equation in the form of a quadratic equation. We can see that we have even powers of π‘₯, meaning we can replace π‘₯ with another variable, say 𝑦, giving us 𝑦=π‘₯. Doing so gives us π‘₯βˆ’10π‘₯+1=0π‘¦βˆ’10𝑦+1=0οŠͺ

We can now see that π‘¦βˆ’10𝑦+1=0 takes the form of a quadratic equation π‘Žπ‘¦+𝑏𝑦+𝑐=0, π‘Žβ‰ 0. We can then apply the quadratic formula to solve for 𝑦, which states 𝑦=βˆ’π‘Β±βˆšπ‘βˆ’4π‘Žπ‘2π‘Ž.

Since π‘Ž=1, 𝑏=βˆ’10, and 𝑐=1, then substituting gives us 𝑦=βˆ’(βˆ’10)±(βˆ’10)βˆ’4(1)(1)2(1).

Simplifying, we then get 𝑦=10±√100βˆ’42=10±√962=10Β±4√62=5Β±2√6.

Remember that we let 𝑦=π‘₯, so π‘₯=5Β±2√6.

Square rooting both sides and solving for π‘₯ gives us π‘₯=±5Β±2√6.

Therefore, all solutions to the equation π‘₯βˆ’10π‘₯+1=0οŠͺ are π‘₯=5+2√6,π‘₯=βˆ’ο„5+2√6,π‘₯=5βˆ’2√6,π‘₯=βˆ’ο„5βˆ’2√6.and

In this explainer, we have discussed how to solve quadratic equations using the quadratic formula. In order to get equations into the form required, we have either rearranged the equation, or substituted to make it into a quadratic form. Let’s recap the key points.

Key Points

  • We can solve quadratic equations of the form π‘Žπ‘₯+𝑏π‘₯+𝑐=0, π‘Žβ‰ 0 using the quadratic formula π‘₯=βˆ’π‘Β±βˆšπ‘βˆ’4π‘Žπ‘2π‘Ž.
  • By rearranging, some equations can be written as a quadratic equations of the form π‘Žπ‘₯+𝑏π‘₯+𝑐=0, π‘Žβ‰ 0, and then solved using the quadratic formula.
  • Some equations that are not quadratic equations themselves, can be solved using the quadratic formula if they take the form π‘Žπ‘₯+𝑏π‘₯+𝑐=0, π‘Žβ‰ 0, where π‘₯ can represent another function.

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