Lesson Explainer: Quadratic Equations: Coefficients and Roots Mathematics

In this explainer, we will learn how to recognize the relationship between the coefficients of a quadratic equation and its roots.

We begin with a brief review of how to factor a quadratic equation to find its roots. Consider the equation π‘₯βˆ’7π‘₯+10=0.

The quadratic expression on the left-hand side of this equation can be factored as (π‘₯βˆ’2)(π‘₯βˆ’5)=0, which leads to the roots π‘₯=2 and π‘₯=5. When we look more closely at this factorization process, we first looked for a pair of numbers that had a product of the constant term 10 and sum to the coefficient βˆ’7. After factoring, this pair of numbers turned out to be the negative of the roots of the quadratic equation. Thinking about this process in reverse, the roots of this equation, π‘₯ and π‘₯, should satisfy (βˆ’π‘₯)+(βˆ’π‘₯)=βˆ’7(βˆ’π‘₯)β‹…(βˆ’π‘₯)=10.and

In other words, we are starting with the presumption that the coefficients of the quadratic equation contain information about the roots. We can verify that this is true in this example, since we know that the roots are π‘₯=2 and π‘₯=5, and (βˆ’2)+(βˆ’5)=βˆ’7(βˆ’2)β‹…(βˆ’5)=10.and

But is this true for any quadratic equation? What is the precise relationship between the coefficients of any quadratic equation and its roots?

Let’s start with a simple quadratic equation, π‘₯+𝑏π‘₯+𝑐=0, whose leading coefficient (i.e., the coefficient of π‘₯) is 1. Say that this equation factors into (π‘₯βˆ’π‘₯)(π‘₯βˆ’π‘₯)=0 for some π‘₯ and π‘₯. Then, the quadratic equation π‘₯+𝑏π‘₯+𝑐=0 has roots π‘₯ and π‘₯. We distribute this equation across the parentheses: π‘₯βˆ’π‘₯π‘₯βˆ’π‘₯π‘₯+π‘₯π‘₯=0.

The middle two terms have π‘₯ in common, so we can take out this common factor to write π‘₯βˆ’(π‘₯+π‘₯)π‘₯+π‘₯π‘₯=0.

After the expansion is complete, we can compare the coefficients of this equation with those of the original equation, π‘₯+𝑏π‘₯+𝑐=0. Comparing this equation with the original quadratic equation, π‘₯+𝑏π‘₯+𝑐=0, we note that 𝑏=βˆ’(π‘₯+π‘₯)𝑐=π‘₯π‘₯.and

In other words, quadratic equations of the form π‘₯+𝑏π‘₯+𝑐=0 with roots π‘₯ and π‘₯ must satisfy π‘₯+π‘₯=βˆ’π‘π‘₯π‘₯=𝑐.and

So, what can we say about the coefficients in a more general quadratic equation, where the leading coefficient is not equal to 1? Consider a quadratic equation π‘Žπ‘₯+𝑏π‘₯+𝑐=0 with roots π‘₯ and π‘₯. Since π‘Žβ‰ 0, We can divide both sides by π‘Ž to write π‘₯+π‘π‘Žπ‘₯+π‘π‘Ž=0.

Now, we have a quadratic equation with a leading coefficient of 1. Then, using our previous results, the coefficients and the roots of the quadratic equation must satisfy the following equations.

Theorem: Coefficients and Roots of a Quadratic Equation

Let π‘Žπ‘₯+𝑏π‘₯+𝑐=0 be a quadratic equation with roots π‘₯ and π‘₯. Then, the roots must satisfy π‘₯+π‘₯=βˆ’π‘π‘Žπ‘₯β‹…π‘₯=π‘π‘Ž.and

For simpler quadratic equations in the form π‘₯+𝑏π‘₯+𝑐=0 with roots π‘₯ and π‘₯, we have the abbreviated formulae π‘₯+π‘₯=βˆ’π‘π‘₯β‹…π‘₯=𝑐.and

These formulae stand true for all quadratic equations, even when the roots are complex valued or are repeated. The same formulae can be recovered using the quadratic formula.

For example, consider the quadratic equation 7π‘₯+2π‘₯+20=0. Without solving the equation, we can find the sum and product of its roots. Since π‘Ž=7,𝑏=2,𝑐=20, we have π‘₯+π‘₯=βˆ’π‘π‘Ž=βˆ’27π‘₯β‹…π‘₯=π‘π‘Ž=207,and where π‘₯ and π‘₯ are the roots of this quadratic equation.

We can verify this by computing the roots using the quadratic roots formula. We have π‘₯=βˆ’π‘Β±βˆšπ‘βˆ’4π‘Žπ‘2π‘Ž=βˆ’2±√2βˆ’4β‹…7β‹…202β‹…7=βˆ’2Β±βˆšβˆ’55614=βˆ’17Β±π‘–βˆš1397.

So, we have π‘₯=βˆ’17+π‘–βˆš1397π‘₯=βˆ’17βˆ’π‘–βˆš1397.and

Then, π‘₯+π‘₯=ο€Ώβˆ’17+π‘–βˆš1397+ο€Ώβˆ’17βˆ’π‘–βˆš1397=βˆ’27, which agrees with what we obtained above using the theorem.

Also, using the difference of squares formula, (π‘₯+𝑦)(π‘₯βˆ’π‘¦)=π‘₯βˆ’π‘¦οŠ¨οŠ¨, we can compute π‘₯β‹…π‘₯=ο€Ώβˆ’17+π‘–βˆš1397ο‹β‹…ο€Ώβˆ’17βˆ’π‘–βˆš1397=ο€Όβˆ’17οˆβˆ’ο€Ώπ‘–βˆš1397=149βˆ’π‘–13949=1+13949=14049=207, which also agrees with our result using the theorem.

In our first example, we will demonstrate how our theorem can help us find the sum of the roots without solving the equation.

Example 1: Relation between the Coefficients of a Quadratic Equation and Its Roots

Without solving the equation βˆ’3π‘₯βˆ’16π‘₯+63=0, find the sum of its roots.

Answer

We recall that, for quadratic equations π‘Žπ‘₯+𝑏π‘₯+𝑐=0 with roots π‘₯ and π‘₯, π‘₯+π‘₯=βˆ’π‘π‘Žπ‘₯β‹…π‘₯=π‘π‘Ž.and

Since the given quadratic equation is βˆ’3π‘₯βˆ’16π‘₯+63=0, we have π‘Ž=βˆ’3,𝑏=βˆ’16,𝑐=63.

Using the formula above, the sum of its roots is equal to π‘₯+π‘₯=βˆ’π‘π‘Ž=βˆ’βˆ’16βˆ’3=βˆ’163.

So, the sum of its roots is equal to βˆ’163.

Using the relationship between the coefficients and the roots of quadratic equations, we can find quadratic equations given their roots. This is the reverse process to problems where we find the roots of a quadratic equation. For these types of problems, we can also verify our answer by finding the roots of the quadratic equation and comparing them with the given roots.

When deriving a quadratic equation from the roots, it is easier to start with the simpler form π‘₯+𝑏π‘₯+𝑐=0 where the leading coefficient is equal to 1. If either of the coefficients 𝑏 and 𝑐 is a fraction, then we can multiply both sides of the equation by the common denominator to further simplify the quadratic equation.

For example, let us derive a quadratic equation with roots 72 and βˆ’2. Starting with the quadratic equation π‘₯+𝑏π‘₯+𝑐=0, we can use our theorem to write 72βˆ’2=βˆ’π‘βŸΉπ‘=βˆ’32 and 72β‹…(βˆ’2)=π‘βŸΉπ‘=βˆ’7.

So, we get 𝑏=βˆ’32 and 𝑐=βˆ’7, leading to the quadratic equation π‘₯βˆ’32π‘₯βˆ’7=0.

Since this equation contains a noninteger coefficient, 32, we can further simplify this equation by multiplying both sides by 2. Then, we get 2π‘₯βˆ’3π‘₯βˆ’14=0.

So, the simplest quadratic equation with roots 72 and βˆ’2 is 2π‘₯βˆ’3π‘₯βˆ’14=0.

We can verify our answer by solving the quadratic equation by factoring. We can factor the quadratic expression 2π‘₯βˆ’3π‘₯βˆ’14 as (2π‘₯βˆ’7)(π‘₯+2), so (2π‘₯βˆ’7)(π‘₯+2)=0.

This leads to the roots π‘₯=72 and π‘₯=βˆ’2, which agrees with the given roots. This verifies our answer.

We will now consider a few examples where we derive the quadratic equation from the roots.

Example 2: The Relation between the Coefficient of a Quadratic Equation and Its Roots

Given that βˆ’1 and βˆ’6 are the solutions of the equation π‘₯+𝑏π‘₯+𝑐=0, find the values of 𝑏 and 𝑐.

Answer

We recall that, for quadratic equations of the form π‘₯+𝑏π‘₯+𝑐=0 with roots π‘₯ and π‘₯, π‘₯+π‘₯=βˆ’π‘π‘₯β‹…π‘₯=𝑐.and

We are given that π‘₯=βˆ’1 and π‘₯=βˆ’6 are roots of the equation, so βˆ’1βˆ’6=βˆ’π‘, which leads to 𝑏=7. Also, we have (βˆ’1)β‹…(βˆ’6)=𝑐, leading to 𝑐=6.

We verify our answer by working backward to solve the quadratic equation by factoring. Our answer leads to the quadratic equation π‘₯+7π‘₯+6=0.

By factoring (π‘₯+1)(π‘₯+6)=0, we can obtain the roots π‘₯=βˆ’1 and π‘₯=βˆ’6. These roots agree with the ones given by the problem.

So, 𝑏=7 and 𝑐=6.

Example 3: Forming Quadratic Equations in the Simplest Form given Their Roots

Find, in its simplest form, the quadratic equation whose roots are βˆ’3 and βˆ’8.

Answer

Since we are looking for the simplest form of a quadratic equation, we should look for a quadratic equation with integer coefficients. Let us start with the form π‘₯+𝑏π‘₯+𝑐=0. If any of the coefficients are fractions, then we can multiply both sides of the equation by the common denominator to simplify the equation.

We remember that if π‘₯ and π‘₯ are the roots of this equation, then π‘₯+π‘₯=βˆ’π‘π‘₯β‹…π‘₯=𝑐.and

We are given that π‘₯=βˆ’3 and π‘₯=βˆ’8 are its roots, so βˆ’3βˆ’8=βˆ’π‘, leading to 𝑏=11. Also, we have (βˆ’3)β‹…(βˆ’8)=𝑐, leading to 𝑐=24. Since the coefficients 𝑏 and 𝑐 are integers, we do not need to alter this equation further. So we obtain the quadratic equation π‘₯+11π‘₯+24=0.

We can verify our answer by solving the quadratic equation by factoring: π‘₯+11π‘₯+24=0⟹(π‘₯+3)(π‘₯+8)=0, so we obtain the roots π‘₯=βˆ’3 and π‘₯=βˆ’8. These roots agree with the ones given in the problem.

So the quadratic equation with roots βˆ’3 and βˆ’8, in its simplest form, is π‘₯+11π‘₯+24=0.

Example 4: Forming a Quadratic Equation in the Simplest Form given Its Roots

Find, in its simplest form, the quadratic equation whose roots are 5+√2 and 5βˆ’βˆš2.

Answer

Let us start with the form π‘₯+𝑏π‘₯+𝑐=0. If any of the coefficients are fractions, then we can multiply both sides of the equation by the common denominator to simplify the equation. We remember that if π‘₯ and π‘₯ are the roots of this equation, then π‘₯+π‘₯=βˆ’π‘π‘₯β‹…π‘₯=𝑐.and

We are given that π‘₯=5+√2 and π‘₯=5βˆ’βˆš2 are its roots, so βˆ’π‘=ο€»5+√2+ο€»5βˆ’βˆš2=(5+5)+ο€»βˆš2βˆ’βˆš2=10, which leads to 𝑏=βˆ’10. Also, recalling the difference of squares formula, (π‘₯+𝑦)(π‘₯βˆ’π‘¦)=π‘₯βˆ’π‘¦οŠ¨οŠ¨, we get 𝑐=ο€»5+√2⋅5βˆ’βˆš2=5βˆ’ο€»βˆš2=25βˆ’2=23.

So we obtain the quadratic equation π‘₯βˆ’10π‘₯+23=0.

We can verify our answer by solving the quadratic equation using the quadratic formula: π‘₯=βˆ’(βˆ’10)±(βˆ’10)βˆ’4β‹…1β‹…232β‹…1=10±√82=10Β±2√22=5±√2.

These roots agree with the ones given in the problem.

So the quadratic equation with roots 5+√2 and 5βˆ’βˆš2, in its simplest form, is π‘₯βˆ’10π‘₯+23=0.

In our next example, we will demonstrate how the procedure holds for complex roots.

Example 5: Identifying Quadratic Equations with a Given Pair of Complex Roots

Which quadratic equation has roots π‘₯=2±𝑖?

  1. π‘₯βˆ’4π‘₯+5=0
  2. π‘₯+4π‘₯+5=0
  3. π‘₯βˆ’4π‘₯+3=0
  4. π‘₯+4π‘₯+3=0
  5. π‘₯βˆ’5π‘₯+4=0

Answer

Since all the given choices carry 1 as the leading coefficient, we start with the quadratic equation π‘₯+𝑏π‘₯+𝑐=0. We remember that if π‘₯ and π‘₯ are the roots of this equation, then π‘₯+π‘₯=βˆ’π‘π‘₯β‹…π‘₯=𝑐.and

We are given that π‘₯=2+π‘–οŠ§ and π‘₯=2βˆ’π‘–οŠ¨ are its roots, so βˆ’π‘=(2+𝑖)+(2βˆ’π‘–)=(2+2)+(π‘–βˆ’π‘–)=4, which leads to 𝑏=βˆ’4. Also, recalling the difference of squares formula, (π‘₯+𝑦)(π‘₯βˆ’π‘¦)=π‘₯βˆ’π‘¦οŠ¨οŠ¨, we get 𝑐=(2+𝑖)β‹…(2βˆ’π‘–)=2βˆ’(𝑖)=4βˆ’(βˆ’1)=5.

We obtain the quadratic equation π‘₯βˆ’4π‘₯+5=0.

We can verify our answer by solving the quadratic equation using the quadratic formula: π‘₯=βˆ’(βˆ’4)±(βˆ’4)βˆ’4β‹…1β‹…52β‹…1=4Β±βˆšβˆ’42=4Β±2𝑖2=2±𝑖.

These roots agree with the ones given in the problem.

So the quadratic equation with roots 2+𝑖 and 2βˆ’π‘–, in its simplest form, is π‘₯βˆ’4π‘₯+5=0.

In the last two examples, we will look at quadratic equations containing an unknown. We will use the relationship between the coefficients and the roots of quadratic equations to solve these problems.

Example 6: Finding the Value of an Unknown in a Quadratic Equation given One of Its Roots

Given that π‘₯=βˆ’9 is a root of the equation π‘₯+π‘šπ‘₯=36, determine the value of π‘š.

Answer

We begin by putting the given equation in the standard form, π‘₯+π‘šπ‘₯βˆ’36=0.

We recall that, given a quadratic equation π‘₯+𝑏π‘₯+𝑐=0, 𝑐 is equal to the product of its roots. In our example, 𝑐=βˆ’36. We are given one of its roots, π‘₯=βˆ’9, but we do not know the other root, which we will denote π‘₯.

Then, βˆ’9β‹…π‘₯=βˆ’36⟹π‘₯=βˆ’36βˆ’9=4.

So the second root of this equation is π‘₯=4.

We also recall that, given a quadratic equation π‘₯+𝑏π‘₯+𝑐=0, βˆ’π‘ is equal to the sum of its roots. In our example, 𝑏=π‘š. Since we know that the roots are π‘₯=βˆ’9 and π‘₯=4, then βˆ’9+4=βˆ’π‘šβŸΉβˆ’5=βˆ’π‘šβŸΉπ‘š=5.

So, π‘š=5.

Example 7: Finding the Value of an Algebraic Expression Using the Relation between the Coefficients of a Quadratic Equation and Its Roots

If 𝐿 and 𝑀 are the roots of the equation π‘₯+10π‘₯+9=0, what is the value of 𝐿+π‘€οŠ¨οŠ¨?

Answer

There are two ways to approach this problem. The first is to use the relation between the coefficients and the roots of a quadratic equation to reproduce the required expression. The second is to find the roots of the quadratic equation and then to compute the required expression. We begin with the first approach, and we will follow with the second one.

We recall that, given a quadratic equation π‘₯+𝑏π‘₯+𝑐=0 with roots 𝐿 and 𝑀, 𝐿+𝑀=βˆ’π‘πΏβ‹…π‘€=𝑐.and

In our example, we note that 𝑏=10 and 𝑐=9, so 𝐿+𝑀=βˆ’10𝐿𝑀=9.and

Now, we observe that the required expression 𝐿+π‘€οŠ¨οŠ¨ can be written as 𝐿+𝑀=𝐿+𝑀+2πΏπ‘€ο…βˆ’2𝐿𝑀=(𝐿+𝑀)βˆ’2𝐿𝑀.

Since we know 𝐿+𝑀=βˆ’10 and 𝐿𝑀=9, 𝐿+𝑀=(βˆ’10)βˆ’2β‹…9=100βˆ’18=82.

So, 𝐿+𝑀=82, following the first method.

Now, let us examine the second method. Given the equation π‘₯+10π‘₯+9=0, we can factor the left-hand side, (π‘₯+1)(π‘₯+9)=0, which gives us the roots π‘₯=βˆ’1 and π‘₯=βˆ’9. We can assign 𝐿=βˆ’1 and 𝑀=βˆ’9 (the result would be the same if we assigned 𝐿=βˆ’9 and 𝑀=βˆ’1, because the required expression is 𝐿+π‘€οŠ¨οŠ¨ and addition is commutative).

We compute the required expression: 𝐿+𝑀=(βˆ’1)+(βˆ’9)=1+81=82.

So 𝐿+π‘€οŠ¨οŠ¨ is equal to 82.

Key Points

  • Coefficients of a quadratic equation contain information about the sum and the product of its roots.
  • For quadratic equations π‘₯+𝑏π‘₯+𝑐=0, if π‘₯ and π‘₯ are the roots, then π‘₯+π‘₯=βˆ’π‘π‘₯β‹…π‘₯=𝑐.and
  • For general quadratic equations π‘Žπ‘₯+𝑏π‘₯+𝑐=0, if π‘₯ and π‘₯ are the roots, then π‘₯+π‘₯=βˆ’π‘π‘Žπ‘₯β‹…π‘₯=π‘π‘Ž.and
  • When deriving a quadratic equation π‘₯+𝑏π‘₯+𝑐=0 from given roots π‘₯ and π‘₯, we can use the abbreviated formulae π‘₯+π‘₯=βˆ’π‘π‘₯β‹…π‘₯=𝑐.and

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