Lesson Explainer: Forming a Quadratic Equation Using Another Quadratic Equation Mathematics

In this explainer, we will learn how to write a quadratic equation given the roots of another quadratic equation.

Consider a quadratic equation in the form π‘Žπ‘₯+𝑏π‘₯+𝑐=0, where π‘Ž, 𝑏, and 𝑐 are constants, and π‘Ž is not equal to 0. The quadratic formula states that the solutions to, or roots of, the equation are π‘₯=βˆ’π‘+βˆšπ‘βˆ’4π‘Žπ‘2π‘Žπ‘₯=βˆ’π‘βˆ’βˆšπ‘βˆ’4π‘Žπ‘2π‘Ž.and

Algebraically, we can show that π‘₯+π‘₯, or the sum of these two roots, is equal to βˆ’π‘π‘Ž, and that π‘₯π‘₯, or the product of these two roots, is equal to π‘π‘Ž, as follows:

This allows us to make a generalization about the relationship between a quadratic equation and its roots.

Property: Relationship between a Quadratic Equation and Its Roots

For any quadratic equation in the form π‘Žπ‘₯+𝑏π‘₯+𝑐=0, the sum of the roots π‘₯ and π‘₯ is equal to βˆ’π‘π‘Ž, and the product of the roots is equal to π‘π‘Ž. That is, π‘₯+π‘₯=βˆ’π‘π‘Žπ‘₯π‘₯=π‘π‘Ž.and

Notice that if π‘Ž, or the leading coefficient in the equation π‘Žπ‘₯+𝑏π‘₯+𝑐=0, is equal to 1, then the sum of the roots becomes π‘₯+π‘₯=βˆ’π‘1=βˆ’π‘, and the product of the roots becomes π‘₯π‘₯=𝑐1=𝑐.

We can, therefore, make another generalization about the relationship between a quadratic equation and its roots when the leading coefficient of the equation is 1.

Property: Relationship between a Quadratic Equation with a Leading Coefficient of 1 and Its Roots

For any quadratic equation in the form π‘Žπ‘₯+𝑏π‘₯+𝑐=0 for which π‘Ž=1, the sum of the roots π‘₯ and π‘₯ is equal to βˆ’π‘, or the negative of the coefficient of π‘₯, and the product of the roots is equal to 𝑐, or the constant term. That is, π‘₯+π‘₯=βˆ’π‘π‘₯π‘₯=𝑐.and

Using these relationships, we are able to form a quadratic equation using another quadratic equation and its roots. For example, if we are given that the roots of the equation π‘₯–7π‘₯+10=0 are 𝐿–1 and 𝑀–1, we know that the sum of these roots is equal to –(–7), or 7 and that the product of these roots is equal to 10. With this knowledge, we can form equations that we can use to help determine the quadratic equation whose roots are 𝐿 and 𝑀. Let’s look at how we would solve problems such as this in the examples that follow.

Example 1: Forming a Quadratic Equation Using Another Quadratic Equation and Its Roots

Given that 𝐿+3 and 𝑀+3 are the roots of the equation π‘₯+8π‘₯+12=0, find, in its simplest form, the quadratic equation whose roots are 𝐿 and 𝑀.

Answer

We know that for any quadratic equation in the form π‘Žπ‘₯+𝑏π‘₯+𝑐=0 for which π‘Ž=1, the sum of the roots is equal to –𝑏, or the negative of the coefficient of π‘₯, and the product of the roots is equal to 𝑐, or the constant term.

In the quadratic equation π‘₯+8π‘₯+12=0, we can see that the value of π‘Ž is 1, the value of 𝑏 is 8, and the value of 𝑐 is 12. This means that since the roots of this equation are 𝐿+3 and 𝑀+3, we know that 𝐿+3+𝑀+3=–8(𝐿+3)(𝑀+3)=12.and

The problem asks us to find another quadratic equation in its simplest form whose roots are 𝐿 and 𝑀. Suppose this other quadratic equation is in the form π‘Žπ‘₯+𝑏π‘₯+𝑐=0. If we assume that π‘Ž=1, then we know that –𝑏 must be equal to 𝐿+𝑀 and that 𝑐 must be equal to 𝐿𝑀. First, let’s use the equation 𝐿+3+𝑀+3=–8 to help find a value for the expression 𝐿+𝑀: 𝐿+3+𝑀+3=–8𝐿+𝑀+6=–8𝐿+𝑀+6–6=–8–6𝐿+𝑀=–14.

Since the value of 𝐿+𝑀 is –14, we know that this is the value of –𝑏, and hence, the value of 𝑏 must be 14.

Next, let’s use the equation (𝐿+3)(𝑀+3)=12 to help find a value for the expression 𝐿𝑀: (𝐿+3)(𝑀+3)=12𝐿𝑀+3𝐿+3𝑀+9=12𝐿𝑀+3𝐿+3𝑀=3𝐿𝑀+3(𝐿+𝑀)=3.

Notice that in the second term on the left side of the equation, we have the expression 𝐿+𝑀 being multiplied by 3. We already know that the value of the expression 𝐿+𝑀 is –14, so we can now substitute –14 for 𝐿+𝑀 and then isolate 𝐿𝑀: 𝐿𝑀+3(–14)=3𝐿𝑀+(–42)=3𝐿𝑀=45.

Since 𝐿𝑀 is equal to 45, we know that this is the value of 𝑐. Therefore, since π‘Ž=1, 𝑏=14, and 𝑐=45, we have shown that the quadratic equation π‘₯+14π‘₯+45=0 has the roots 𝐿 and 𝑀.

Now, let’s look at a similar example in which we must form a quadratic equation using another quadratic equation and its roots.

Example 2: Forming a Quadratic Equation Using Another Quadratic Equation and Its Roots

Given that 𝐿 and 𝑀 are the roots of the equation π‘₯–2π‘₯+5=0, find, in its simplest form, the quadratic equation whose roots are 𝐿 and π‘€οŠ¨.

Answer

Recall that for any quadratic equation in the form π‘Žπ‘₯+𝑏π‘₯+𝑐=0 for which π‘Ž=1, the sum of the roots is equal to –𝑏, or the negative of the coefficient of π‘₯, and the product of the roots is equal to 𝑐, or the constant term.

We can see that, in the quadratic equation π‘₯–2π‘₯+5=0, the value of π‘Ž is 1, the value of 𝑏 is –2, and the value of 𝑐 is 5. We were told that the roots of this equation are 𝐿 and 𝑀, so we know that 𝐿+𝑀=–(–2)=2𝐿𝑀=5.and

The problem asks us to find another quadratic equation in its simplest form whose roots are 𝐿 and π‘€οŠ¨. Suppose this other quadratic equation is in the form π‘Žπ‘₯+𝑏π‘₯+𝑐=0. If we assume that π‘Ž=1, then we know that βˆ’π‘ must be equal to 𝐿+π‘€οŠ¨οŠ¨ and that 𝑐 must be equal to πΏπ‘€οŠ¨οŠ¨. In order to find a value for the expression 𝐿+π‘€οŠ¨οŠ¨, we must begin by squaring both sides of the equation 𝐿+𝑀=2 and then expanding the left side: (𝐿+𝑀)=2(𝐿+𝑀)(𝐿+𝑀)=4𝐿+𝐿𝑀+𝐿𝑀+𝑀=4𝐿+2𝐿𝑀+𝑀=4.

Notice that in the second term on the left side of the equation, we have the expression 𝐿𝑀 being multiplied by 2. Since we already know that the value of the expression 𝐿𝑀 is 5, we can now substitute 5 for 𝐿𝑀 and then isolate 𝐿+π‘€οŠ¨οŠ¨: 𝐿+2(5)+𝑀=4𝐿+10+𝑀=4𝐿+𝑀=–6.

Since the value of 𝐿+π‘€οŠ¨οŠ¨ is –6, we know that this is the value of –𝑏, and hence, the value of 𝑏 must be 6.

Now, let’s find a value for the expression πΏπ‘€οŠ¨οŠ¨. We can do so by squaring both sides of the equation 𝐿𝑀=5 and then distributing the exponent:𝐿𝑀=5(𝐿𝑀)=5𝐿𝑀=25.

Since πΏπ‘€οŠ¨οŠ¨ is equal to 25, we know that this is the value of 𝑐. Therefore, since π‘Ž=1, 𝑏=6, and 𝑐=25, we have shown that the quadratic equation π‘₯+6π‘₯+25=0 has the roots 𝐿 and π‘€οŠ¨.

In the problem that follows, we will find the value of an expression using the relationship between the coefficients of a quadratic equation and its roots.

Example 3: Using the Relationship between the Coefficients of a Quadratic Equation and Its Roots to Find the Value of an Expression

If 𝐿 and 𝑀 are the roots of the equation π‘₯+20π‘₯+15=0, what is the value of 1𝑀+1𝐿?

Answer

The equation π‘₯+20π‘₯+15=0 is quadratic and is in the form π‘Žπ‘₯+𝑏π‘₯+𝑐=0, where π‘Ž=1, 𝑏=20, and 𝑐=15. Since the value of π‘Ž is 1, we know that the sum of the equation’s roots is equal to –𝑏, or the negative of the coefficient of π‘₯, and that the product of its roots is equal to 𝑐, or the constant term. We are told that the equation’s roots are 𝐿 and 𝑀, so we can therefore write the equations 𝐿+𝑀=–20𝐿𝑀=15.and

The problem asks us to find the value of 1𝑀+1𝐿. In order to do so, we must rewrite the two fractions in the expression with a common denominator and then add the fractions together. We can begin by multiplying the numerator and denominator of each fraction by the denominator of the other to get 1⋅𝐿𝑀⋅𝐿+1⋅𝑀𝐿⋅𝑀. Simplifying then gives us 𝐿𝐿𝑀+𝑀𝐿𝑀, and after adding the fractions together, we get 𝐿+𝑀𝐿𝑀.

Notice that the numerator of the resulting expression is the sum of the roots of the equation π‘₯+20π‘₯+15=0 and that the denominator of the expression is the product of the roots. We have already determined that the sum of the roots, or 𝐿+𝑀, is equal to –20, and that the product of the roots, or 𝐿𝑀, is equal to 15, so the value of 1𝑀+1𝐿 must be βˆ’2015, or βˆ’43.

Next, we will again form a quadratic equation using another quadratic equation and its roots. However, this time, the coefficients we arrive at will not be integers, so we will have to multiply both sides of the equation by a constant to eliminate the fractions.

Example 4: Forming a Quadratic Equation Using Another Quadratic Equation and Its Roots

Given that 𝐿 and 𝑀 are the roots of the equation π‘₯–3π‘₯+12=0, find, in its simplest form, the quadratic equation whose roots are 1𝐿 and 1π‘€οŠ¨.

Answer

Remember that for any quadratic equation in the form π‘Žπ‘₯+𝑏π‘₯+𝑐=0 for which π‘Ž=1, the sum of the roots is equal to βˆ’π‘, or the negative of the coefficient of π‘₯, and the product of the roots is equal to 𝑐, or the constant term.

In the quadratic equation π‘₯–3π‘₯+12=0, we can see that the value of π‘Ž is 1, the value of 𝑏 is βˆ’3, and the value of 𝑐 is 12. Thus, since the roots of this equation are 𝐿 and 𝑀, we know that 𝐿+𝑀=–(–3)=3𝐿𝑀=12.and

The problem asks us to find another quadratic equation in its simplest form whose roots are 1𝐿 and 1π‘€οŠ¨. Suppose this other quadratic equation is in the form π‘Žπ‘₯+𝑏π‘₯+𝑐=0. If we assume that π‘Ž=1, then we know that βˆ’π‘ must be equal to 1𝐿+1π‘€οŠ¨οŠ¨ and that 𝑐 must be equal to 1πΏπ‘€οŠ¨οŠ¨.

In order to find a value for the expression 1𝐿+1π‘€οŠ¨οŠ¨, we must begin by finding a common denominator for the two fractions in the expression and then adding the fractions together to get 𝑀+πΏπΏπ‘€οŠ¨οŠ¨οŠ¨οŠ¨, or 𝐿+π‘€πΏπ‘€οŠ¨οŠ¨οŠ¨οŠ¨.

Now, let’s square both sides of the equation 𝐿+𝑀=3 and then expand the left side: (𝐿+𝑀)=3(𝐿+𝑀)(𝐿+𝑀)=9𝐿+𝐿𝑀+𝐿𝑀+𝑀=9𝐿+2𝐿𝑀+𝑀=9.

Notice that in the second term on the left side of the equation, we have the expression 𝐿𝑀 being multiplied by 2. Since we already know that the value of the expression 𝐿𝑀 is 12, we can substitute 12 for 𝐿𝑀 and then isolate 𝐿+π‘€οŠ¨οŠ¨: 𝐿+2(12)+𝑀=9𝐿+24+𝑀=9𝐿+𝑀=–15.

Next, let’s find a value for the expression πΏπ‘€οŠ¨οŠ¨. We can do so by squaring both sides of the equation 𝐿𝑀=12 and then distributing the exponent: 𝐿𝑀=12(𝐿𝑀)=12𝐿𝑀=144.

Now, substituting 𝐿+𝑀=–15 and 𝐿𝑀=144 into the expression 𝐿+π‘€πΏπ‘€οŠ¨οŠ¨οŠ¨οŠ¨, we get βˆ’15144 for βˆ’π‘, or 15144 for 𝑏, the coefficient of π‘₯ in the equation we are seeking.

Also, substituting 𝐿𝑀=144 into the expression 1πΏπ‘€οŠ¨οŠ¨, we get 1144 for 𝑐, the constant term in the equation. This gives us an equation of π‘₯+15144π‘₯+1144=0.

Finally, to eliminate the fractions, we can multiply both sides by 144 to get 144ο€Όπ‘₯+15144π‘₯+1144=144(0), or 144π‘₯+15π‘₯+1=0, for our equation.

Now, let’s look at an example of how to form a quadratic equation using a nonmonic quadratic equation and its roots. A nonmonic quadratic equation has a leading coefficient that is not 1.

Example 5: Forming a Quadratic Equation Using a Nonmonic Quadratic Equation and Its Roots

Given that 𝐿 and 𝑀 are the roots of the equation 3π‘₯+16π‘₯–1=0, find, in its simplest form, the quadratic equation whose roots are 𝐿2 and 𝑀2.

Answer

We know that for any quadratic equation in the form π‘Žπ‘₯+𝑏π‘₯+𝑐=0, the sum of the roots is equal to βˆ’π‘π‘Ž, and the product of the roots is equal to π‘π‘Ž. In the equation 3π‘₯+16π‘₯–1=0, the value of π‘Ž is 3, the value of 𝑏 is 16, and the value of 𝑐 is –1. Thus, since the roots of this equation are 𝐿 and 𝑀, we know that 𝐿+𝑀=βˆ’163𝐿𝑀=βˆ’13.and

The problem asks us to find another quadratic equation in its simplest form whose roots are 𝐿2 and 𝑀2. Suppose this other quadratic equation is in the form π‘Žπ‘₯+𝑏π‘₯+𝑐=0. Remember that for any quadratic equation in the form π‘Žπ‘₯+𝑏π‘₯+𝑐=0 for which π‘Ž=1, the sum of the roots is equal to βˆ’π‘, or the negative of the coefficient of π‘₯, and the product of the roots is equal to 𝑐, or the constant term.

Thus, if we assume that π‘Ž=1, then we know that βˆ’π‘ must be equal to 𝐿2+𝑀2, or 𝐿+𝑀2 and that 𝑐 must be equal to 𝐿2𝑀2, or 𝐿𝑀4.

Substituting 𝐿+𝑀=βˆ’163 into the expression 𝐿+𝑀2, we get 2=βˆ’83 for βˆ’π‘, or 83 for 𝑏, the coefficient of π‘₯ in the equation we are seeking.

Also, substituting 𝐿𝑀=βˆ’13 into the expression 𝐿𝑀4, we get 4=βˆ’112 for 𝑐, the constant term in the equation. This gives us an equation of π‘₯+83π‘₯βˆ’112=0.

Finally, to eliminate the fractions, we can multiply both sides by 12 to get 12ο€Όπ‘₯+83π‘₯βˆ’112=12(0), or 12π‘₯+32π‘₯βˆ’1=0, for our equation.

Finally, we will work on another problem in which we must form a quadratic equation using a nonmonic quadratic equation and its roots.

Example 6: Forming Quadratic Equations in the Simplest Form Using the Relation between a Quadratic Equation and Its Roots

If 𝐿 and 𝑀 are the roots of the equation 2π‘₯–3π‘₯+1=0, find, in its simplest form, the quadratic equation whose roots are 2𝐿 and 2π‘€οŠ¨.

Answer

We know that for any quadratic equation in the form π‘Žπ‘₯+𝑏π‘₯+𝑐=0, the sum of the roots is equal to βˆ’π‘π‘Ž, and the product of the roots is equal to π‘π‘Ž. In the equation 2π‘₯–3π‘₯+1=0, the value of π‘Ž is 2, the value of 𝑏 is βˆ’3, and the value of 𝑐 is 1. Thus, since the roots of this equation are 𝐿 and 𝑀, we know that 𝐿+𝑀=βˆ’ο€Όβˆ’32=32𝐿𝑀=12.and

The problem asks us to find another quadratic equation in its simplest form whose roots are 2𝐿 and 2π‘€οŠ¨. Suppose this other quadratic equation is in the form π‘Žπ‘₯+𝑏π‘₯+𝑐=0. Recall that, for any quadratic equation in the form π‘Žπ‘₯+𝑏π‘₯+𝑐=0 for which π‘Ž=1, the sum of the roots is equal to βˆ’π‘, or the negative of the coefficient of π‘₯, and the product of the roots is equal to 𝑐, or the constant term.

Therefore, if we assume that π‘Ž=1, then we know that βˆ’π‘ must be equal to 2𝐿+2π‘€οŠ¨οŠ¨ and that 𝑐 must be equal to 2𝐿2π‘€ο…οŠ¨οŠ¨, or 4πΏπ‘€οŠ¨οŠ¨.

In order to determine a value for the expression 2𝐿+2π‘€οŠ¨οŠ¨, let’s start by squaring both sides of the equation 𝐿+𝑀=32 and then expanding the left side: (𝐿+𝑀)=ο€Ό32(𝐿+𝑀)(𝐿+𝑀)=94𝐿+𝐿𝑀+𝐿𝑀+𝑀=94𝐿+2𝐿𝑀+𝑀=94.

Notice that in the second term on the left side of the equation, we have the expression 𝐿𝑀 being multiplied by 2. Since we already know that the value of the expression 𝐿𝑀 is 12, we can now substitute 12 for 𝐿𝑀 and then isolate 𝐿+π‘€οŠ¨οŠ¨: 𝐿+2ο€Ό12+𝑀=94𝐿+1+𝑀=94𝐿+44+𝑀=94𝐿+𝑀=54.

Next, multiplying both sides of the equation 𝐿+𝑀=54 by 2, we get 2𝐿+𝑀=2ο€Ό542𝐿+2𝑀=52.

Since the value of 2𝐿+2π‘€οŠ¨οŠ¨ is 52, we know that this is the value of βˆ’π‘, and hence, the value of 𝑏 must be βˆ’52.

Now, let’s determine a value for the expression 4πΏπ‘€οŠ¨οŠ¨. We can begin by squaring both sides of the equation 𝐿𝑀=12 and then distributing the exponent: 𝐿𝑀=12(𝐿𝑀)=ο€Ό12οˆπΏπ‘€=14.

Next, multiplying both sides of the equation 𝐿𝑀=14 by 4, we get 4𝐿𝑀=4ο€Ό144𝐿𝑀=1.

Since the value of 4πΏπ‘€οŠ¨οŠ¨ is 1, we know that this is the value of 𝑐.

Thus, we can now substitute π‘Ž=1, 𝑏=βˆ’52, and 𝑐=1 into π‘Žπ‘₯+𝑏π‘₯+𝑐=0 to get the equation π‘₯βˆ’52π‘₯+1=0. Finally, to eliminate the fraction, we can multiply both sides by 2 to get 2ο€Όπ‘₯βˆ’52π‘₯+1=2(0), or 2π‘₯βˆ’5π‘₯+2=0, for our equation.

Now let’s finish by recapping some key points.

Key Points

  • For any quadratic equation in the form π‘Žπ‘₯+𝑏π‘₯+𝑐=0, the sum of the roots is equal to βˆ’π‘π‘Ž, and the product of the roots is equal to π‘π‘Ž.
  • For any quadratic equation in the form π‘Žπ‘₯+𝑏π‘₯+𝑐=0 for which π‘Ž=1, the sum of the roots is equal to βˆ’π‘, or the negative of the coefficient of π‘₯, and the product of the roots is equal to 𝑐, or the constant term.
  • It is possible to form a quadratic equation using another quadratic equation and its roots.
  • A nonmonic quadratic equation is a quadratic equation with a leading coefficient that is not 1.

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