For a student in High school, the major understanding of quadratic equation is to solve a given polynomial with a power of two and find its two respective roots. However, there is actually more to the quadratic equation than just solving for roots. The known roots of a quadratic equation can be actually used to derive the equation itself. In fact, it is possible to find an unknown quadratic equation based on its unknown roots as far as another quadratic equation is related to it.

Using the basic understanding and methods of solving quadratic equations will not be able to work for this type of computation. The only viable way students can solve the kind of problem in our above paragraph is to exploit the relationship between the roots and coefficients of the quadratic equation. The coefficients of a quadratic equation are well related to its roots, and this piece will reveal exactly how this is possible. The formulas that will be derived in this relationship are very useful in using the roots to find a quadratic equation

**What are the Coefficients?**

The coefficients are the constants in a general quadratic equation. Now the quadratic equation is given as

ax^{2} + bx + c = 0

where a, b and c are coefficients.

For example, if 2x^{2} + 10x +12 c = 0 is a quadratic equation, then a = 2, b = 10 and c = 12

**What are the Roots?**

The roots are the solutions of a quadratic equation. After solving a quadratic equation, you will end up with two answers. The answers are what are known as roots or solutions of the equation. The nature of roots can be real or imaginary.

**The Relation Between Coefficients and Roots**

We will try to show how the quadratic coefficients relate to the roots.

**The First Stage of Establishing the Relationship Between Coefficients and Roots**

The quadratic equation is generally given in the form

ax^{2} + bx + c = 0 …… (i)

the above equation will give us two solutions or roots that we will identify as y and z

now the a, b and c are the respective equations of the quadratic equation, and by principle, they have a very high relationship with the roots y and z, and we will show this.

Now, if you have checked out our topics on how to find the quadratic equations using its roots you will know that the solutions of the quadratic equation are usually expressed as

X = y, x = z

Which is gotten from the steps

(x – y)(x – z) = 0

Now if we expand the above we will have

X^{2} – xz –xy + yz = 0

X^{2} – (z + y)x + yz = 0……. (ii)

The (ii) therefore means that

X^{2} – (the sum of the roots)x + (products of the roots) = 0 …..(iii)

Now this formula will always work if you have the roots and want to find the equation.

We will consider a quick example to show this

**Example 1**

Solve the quadratic equation X^{2} – 5x + 6 = 0 and find its roots using the general formula. In the reverse case, use the roots to find the quadratic equation.

**Solution**

X^{2} – 5x + 6 = 0

The general formula for quadratic equation is

X = 3, x = 2

We can use the roots to find the equation as well.

Using the formula in eqn (iii) we will be able to find the quadratic equation

X^{2} – (2 + 3)x + (2 x 3) = 0

X^{2} – (5)x + (6) = 0

X^{2} – 5x + 6 = 0

So we can see that the formula outlined in (iii) can be used to find a quadratic equation when the roots are given.

This example is the first proof of the roots and coefficients. How is this possible? We will explain.

Now, if we look at our quadratic equation in the eye of (i) and (ii), which is the same as (iii), we will see that

x^{2} – 5x + 6 = 0 and X^{2} – (2 + 3)x + 2 x 3 = 0 are very similar

- the b which is given as (– 5) in the initial quadratic equation is actually the difference of the sum of the roots – (2 + 3)
- the c, which is given as 6 in the quadratic equation, is the product of the two roots, which is given as (2 x 3)

We have been able to establish that there is a relationship between the coefficient and the roots. However, this is just one part.

**Second Stage of Establishing the Relationship Between Coefficients and Roots**

Now if we look at the equation (i) and (ii), we will see that

ax^{2} + bx + c = 0 and X^{2} – (z + y)x + yz = 0 are very similar

The b, which is given as in (i) is the – (z + y) in (ii), while the c in (i) is the yz in (ii). The only major difference is the fact that a in (i) does not have a variable in (ii) to match it. This will not be a problem if the coefficient of a in (i) is 1

If a is not 1 in (i) then we will need to find another way to relate the root with the coefficient.

Let us divide eqn (i) with a all through

ax^{2}/a + bx/a + c/a = 0

x^{2 }+ (b/a)x + c/a = 0 …. (iv)

now (iv) which is given as x^{2 }+ (b/a)x + c/a = 0 properly matches (ii) X^{2} – (z + y)x + yz = 0 with

z + y = – b/a …… (v)

yz = c/a ….. (vi)

we have now been able to establish the relationship between roots and coefficients in their entirety. We will now list out how the roots and coefficients are related so that you can always use them.

**X**^{2}– (z + y)x + yz = o is the same as ax^{2}+ bx + c = 0 If a = 1 and as such

z + y = b

yz = c

**If a = a value different from 1,**

z + y = – b/a …… (v)

yz = c/a ….. (vi)

with the above established, we will consider an example to show that the two is different.

**Example 2**

If a Quadratic equation is x^{2} – 3x – 5 = 0 with roots y and z, then find the quadratic equation with roots y^{2} and z^{2}

**Solution**

Now we can find the answer with the relationship between coefficients and determinants.

**Step 1: Determine the objective of the problem and the solution needed**

Our aim is to find the quadratic equation of the roots y^{2 }and z^{2}

The quadratic equation we are looking for we be in the form

X^{2} – (y^{2} + z^{2})x + (y^{2}z^{2}) = 0

It is important not to forget this as it will guide our calculations.

So our aim is to find

y^{2} + z^{2} and y^{2}z^{2} = 0

**Step 2: write out all the variables and what they represent**

a = 1. b = -3, c = – 5

also, y, z, y^{2} and z^{2} are all unknown

**Step 3: find y ^{2}z^{2}**

Recall that;

yz = c/a ….. (vi)

as such

yz = – 5/1 = -5

since yz = 5,

(yz)^{2} = y^{2}z^{2} = (- 5)^{2} = 25

That is mission accomplished.

**Step 4: find y ^{2} + z^{2}**

Y + z = – b/a …… (v)

as such

y + z = – (-3)/1 = 3

now following the answer y + z = 3,

(y + z)^{2} = 3^{2} = 9

It is important to also note that

(z + y)^{2} = y^{2} + 2yz + z^{2}

9 = y^{2} + 2 (-5) + z^{2}

9 = y^{2} – 10 + z^{2}

y^{2} + z^{2} = 19

so we have achieved our aim.

**Step 5:** **Re arrange into the desired quadratic equation.**

y^{2} + z^{2} and y^{2}z^{2} and we have done so.

y^{2} + z^{2} = 19

y^{2}z^{2} = 25

We will rearrange into the unknown quadratic equation we need to find

X^{2} – (y^{2} + z^{2})x + (y^{2}z^{2}) = 0

Implies

X^{2} – 19x + 25 = 0

So we have solved the problem.

Students can go further by using the general formula to find y and z as an assignment.

**Conclusion**

The relationship between coefficients and roots is very important in helping people know exactly how the quadratic equation and its solutions work. The basic understanding of quadratic equations gives a glimpse of how coefficients and roots are related. However, not much is shown. This piece has highlighted this relationship and how it can help students derive quadratic equations with no known roots. Questions that may require a good grasp of coefficients and roots often appear in exams and tests, and this piece has comprehensively handled the topic to help students pass well.