Lesson Explainer: Quadratic Equations: Coefficients and Roots | Nagwa Lesson Explainer: Quadratic Equations: Coefficients and Roots | Nagwa

Lesson Explainer: Quadratic Equations: Coefficients and Roots Mathematics • First Year of Secondary School

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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 𝑥7𝑥+10=0.

The quadratic expression on the left-hand side of this equation can be factored as (𝑥2)(𝑥5)=0, which leads to the roots 𝑥=2 and 𝑥=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 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, 𝑥 and 𝑥 (sometimes referred to as 𝐿 and 𝑀), should satisfy (𝑥)+(𝑥)=7(𝑥)(𝑥)=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 𝑥=2 and 𝑥=5, and (2)+(5)=7(2)(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, 𝑥+𝑏𝑥+𝑐=0, whose leading coefficient (i.e., the coefficient of 𝑥) is 1. Say that this equation factors into (𝑥𝑥)(𝑥𝑥)=0 for some 𝑥 and 𝑥. Then, the quadratic equation 𝑥+𝑏𝑥+𝑐=0 has roots 𝑥 and 𝑥. We distribute this equation across the parentheses: 𝑥𝑥𝑥𝑥𝑥+𝑥𝑥=0.

The middle two terms have 𝑥 in common, so we can take out this common factor to write 𝑥(𝑥+𝑥)𝑥+𝑥𝑥=0.

After the expansion is complete, we can compare the coefficients of this equation with those of the original equation, 𝑥+𝑏𝑥+𝑐=0. Comparing this equation with the original quadratic equation, 𝑥+𝑏𝑥+𝑐=0, we note that 𝑏=(𝑥+𝑥)𝑐=𝑥𝑥.and

In other words, quadratic equations of the form 𝑥+𝑏𝑥+𝑐=0 with roots 𝑥 and 𝑥 must satisfy 𝑥+𝑥=𝑏𝑥𝑥=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 𝑎𝑥+𝑏𝑥+𝑐=0 with roots 𝑥 and 𝑥. Since 𝑎0, We can divide both sides by 𝑎 to write 𝑥+𝑏𝑎𝑥+𝑐𝑎=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 𝑎𝑥+𝑏𝑥+𝑐=0 be a quadratic equation with roots 𝑥 and 𝑥. Then, the roots must satisfy 𝑥+𝑥=𝑏𝑎𝑥𝑥=𝑐𝑎.and

For simpler quadratic equations in the form 𝑥+𝑏𝑥+𝑐=0 with roots 𝑥 and 𝑥, we have the abbreviated formulae 𝑥+𝑥=𝑏𝑥𝑥=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 7𝑥+2𝑥+20=0. Without solving the equation, we can find the sum and product of its roots. Since 𝑎=7,𝑏=2,𝑐=20, we have 𝑥+𝑥=𝑏𝑎=27𝑥𝑥=𝑐𝑎=207,and where 𝑥 and 𝑥 are the roots of this quadratic equation.

We can verify this by computing the roots using the quadratic roots formula. We have 𝑥=𝑏±𝑏4𝑎𝑐2𝑎=2±2472027=2±55614=17±𝑖1397.

So, we have 𝑥=17+𝑖1397𝑥=17𝑖1397.and

Then, 𝑥+𝑥=17+𝑖1397+17𝑖1397=27, which agrees with what we obtained above using the theorem.

Also, using the difference of squares formula, (𝑥+𝑦)(𝑥𝑦)=𝑥𝑦, we can compute 𝑥𝑥=17+𝑖139717𝑖1397=17𝑖1397=149𝑖13949=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 3𝑥16𝑥+63=0, find the sum of its roots.

Answer

We recall that, for quadratic equations 𝑎𝑥+𝑏𝑥+𝑐=0 with roots 𝑥 and 𝑥, 𝑥+𝑥=𝑏𝑎𝑥𝑥=𝑐𝑎.and

Since the given quadratic equation is 3𝑥16𝑥+63=0, we have 𝑎=3,𝑏=16,𝑐=63.

Using the formula above, the sum of its roots is equal to 𝑥+𝑥=𝑏𝑎=163=163.

So, the sum of its roots is equal to 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 𝑥+𝑏𝑥+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 2. Starting with the quadratic equation 𝑥+𝑏𝑥+𝑐=0, we can use our theorem to write 722=𝑏𝑏=32 and 72(2)=𝑐𝑐=7.

So, we get 𝑏=32 and 𝑐=7, leading to the quadratic equation 𝑥32𝑥7=0.

Since this equation contains a noninteger coefficient, 32, we can further simplify this equation by multiplying both sides by 2. Then, we get 2𝑥3𝑥14=0.

So, the simplest quadratic equation with roots 72 and 2 is 2𝑥3𝑥14=0.

We can verify our answer by solving the quadratic equation by factoring. We can factor the quadratic expression 2𝑥3𝑥14 as (2𝑥7)(𝑥+2), so (2𝑥7)(𝑥+2)=0.

This leads to the roots 𝑥=72 and 𝑥=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 1 and 6 are the solutions of the equation 𝑥+𝑏𝑥+𝑐=0, find the values of 𝑏 and 𝑐.

Answer

We recall that, for quadratic equations of the form 𝑥+𝑏𝑥+𝑐=0 with roots 𝑥 and 𝑥, 𝑥+𝑥=𝑏𝑥𝑥=𝑐.and

We are given that 𝑥=1 and 𝑥=6 are roots of the equation, so 16=𝑏, which leads to 𝑏=7. Also, we have (1)(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 𝑥+7𝑥+6=0.

By factoring (𝑥+1)(𝑥+6)=0, we can obtain the roots 𝑥=1 and 𝑥=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 3 and 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 𝑥+𝑏𝑥+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 𝑥 and 𝑥 are the roots of this equation, then 𝑥+𝑥=𝑏𝑥𝑥=𝑐.and

We are given that 𝑥=3 and 𝑥=8 are its roots, so 38=𝑏, leading to 𝑏=11. Also, we have (3)(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 𝑥+11𝑥+24=0.

We can verify our answer by solving the quadratic equation by factoring: 𝑥+11𝑥+24=0(𝑥+3)(𝑥+8)=0, so we obtain the roots 𝑥=3 and 𝑥=8. These roots agree with the ones given in the problem.

So the quadratic equation with roots 3 and 8, in its simplest form, is 𝑥+11𝑥+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 52.

Answer

Let us start with the form 𝑥+𝑏𝑥+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 𝑥 and 𝑥 are the roots of this equation, then 𝑥+𝑥=𝑏𝑥𝑥=𝑐.and

We are given that 𝑥=5+2 and 𝑥=52 are its roots, so 𝑏=5+2+52=(5+5)+22=10, which leads to 𝑏=10. Also, recalling the difference of squares formula, (𝑥+𝑦)(𝑥𝑦)=𝑥𝑦, we get 𝑐=5+252=52=252=23.

So we obtain the quadratic equation 𝑥10𝑥+23=0.

We can verify our answer by solving the quadratic equation using the quadratic formula: 𝑥=(10)±(10)412321=10±82=10±222=5±2.

These roots agree with the ones given in the problem.

So the quadratic equation with roots 5+2 and 52, in its simplest form, is 𝑥10𝑥+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 𝑥=2±𝑖?

  1. 𝑥4𝑥+5=0
  2. 𝑥+4𝑥+5=0
  3. 𝑥4𝑥+3=0
  4. 𝑥+4𝑥+3=0
  5. 𝑥5𝑥+4=0

Answer

Since all the given choices carry 1 as the leading coefficient, we start with the quadratic equation 𝑥+𝑏𝑥+𝑐=0. We remember that if 𝑥 and 𝑥 are the roots of this equation, then 𝑥+𝑥=𝑏𝑥𝑥=𝑐.and

We are given that 𝑥=2+𝑖 and 𝑥=2𝑖 are its roots, so 𝑏=(2+𝑖)+(2𝑖)=(2+2)+(𝑖𝑖)=4, which leads to 𝑏=4. Also, recalling the difference of squares formula, (𝑥+𝑦)(𝑥𝑦)=𝑥𝑦, we get 𝑐=(2+𝑖)(2𝑖)=2(𝑖)=4(1)=5.

We obtain the quadratic equation 𝑥4𝑥+5=0.

We can verify our answer by solving the quadratic equation using the quadratic formula: 𝑥=(4)±(4)41521=4±42=4±2𝑖2=2±𝑖.

These roots agree with the ones given in the problem.

So the quadratic equation with roots 2+𝑖 and 2𝑖, in its simplest form, is 𝑥4𝑥+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 𝑥=9 is a root of the equation 𝑥+𝑚𝑥=36, determine the value of 𝑚.

Answer

We begin by putting the given equation in the standard form, 𝑥+𝑚𝑥36=0.

We recall that, given a quadratic equation 𝑥+𝑏𝑥+𝑐=0, 𝑐 is equal to the product of its roots. In our example, 𝑐=36. We are given one of its roots, 𝑥=9, but we do not know the other root, which we will denote 𝑥.

Then, 9𝑥=36𝑥=369=4.

So the second root of this equation is 𝑥=4.

We also recall that, given a quadratic equation 𝑥+𝑏𝑥+𝑐=0, 𝑏 is equal to the sum of its roots. In our example, 𝑏=𝑚. Since we know that the roots are 𝑥=9 and 𝑥=4, then 9+4=𝑚5=𝑚𝑚=5.

So, 𝑚=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 𝑥+10𝑥+9=0, what is the value of 𝐿+𝑀?

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 𝑥+𝑏𝑥+𝑐=0 with roots 𝐿 and 𝑀, 𝐿+𝑀=𝑏𝐿𝑀=𝑐.and

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

Now, we observe that the required expression 𝐿+𝑀 can be written as 𝐿+𝑀=𝐿+𝑀+2𝐿𝑀2𝐿𝑀=(𝐿+𝑀)2𝐿𝑀.

Since we know 𝐿+𝑀=10 and 𝐿𝑀=9, 𝐿+𝑀=(10)29=10018=82.

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

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

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

So 𝐿+𝑀 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 𝑥+𝑏𝑥+𝑐=0, if 𝑥 and 𝑥 are the roots, then 𝑥+𝑥=𝑏𝑥𝑥=𝑐.and
  • For general quadratic equations 𝑎𝑥+𝑏𝑥+𝑐=0, if 𝑥 and 𝑥 are the roots, then 𝑥+𝑥=𝑏𝑎𝑥𝑥=𝑐𝑎.and
  • When deriving a quadratic equation 𝑥+𝑏𝑥+𝑐=0 from given roots 𝑥 and 𝑥, we can use the abbreviated formulae 𝑥+𝑥=𝑏𝑥𝑥=𝑐.and

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