How to find the Equation of a Quadratic Function?

A quadratic equation is defined as an algebraic equation second-degree equation with one unknown variable. The word quadratic is derived from the Latin word “quadratum” which means “square”.

The standard form of a quadratic equation is given as ax² + bx + c = 0, where a, b are the coefficients and a≠0, c is a constant and x is the unknown variable.

ax² + bx + c = 0

• where a ≠ 0, a, b, & c are Real numbers

Since the highest degree of the variable is two, a quadratic equation always has two roots. The roots of the quadratic equation are defined as the values of x satisfying the quadratic equation where it is equal to zero. And the roots are found by using the quadratic formula,

x = [-b ± √(b² – 4ac)]/2a

Let us consider the general form of the quadratic equation: 

ax² + bx + c = 0, 

• where a ≠ 0 and a, b, c ∈ R.

or

ax² + bx = -c

Now divide both sides with a as,

x² + (b/a)x = -c/a

Now add both sides with the term (b/2a)²

 x² + (b/a)x + (b/2a)2 = -c/a + (b/2a)²

or

x² + 2. (b/2a) x + (b/2a)² = -c/a + (b/2a)²

or

(x + b/2a)² = -c/a + b²/4a²

= (b² – 4ac)/4a²

Now apply square root on both sides as,

x + b/2a = ± (√b² – 4ac)/2a

x = [±(√b² – 4ac)/2a] – b/2a

= (-b ± √b² – 4ac)/2a

So, the roots of the quadratic equation of the quadratic equation ax² + bx + c = 0 are: 

(-b ± √b² – 4ac)/2a

where, 

b² – 4ac is called as the discriminant (D). 

Discriminant determines the nature of the roots i.e., whether the quadratic equation has two real solutions or one real solution, or two complex solutions.

	Discriminant value (D)	Nature of the roots
Case 1	b2 – 4ac > 0	Two real solutions
Case 2	b2 – 4ac = 0	One real solution
Case 3	b2 – 4ac < 0	Two complex solutions

Relation between the Roots and Coefficients of a quadratic equation:

Let α and β be the roots of the quadratic equation which are obtained by solving ax² + bx + c = 0.

• The sum of roots of the quadratic equation = α + β = -b/a = – coefficient of x/ coefficient of x²
• Product of roots of the quadratic equation = αβ = c/a = constant/coefficient of x²
• The quadratic equation having roots α and β is represented as, x² – (α + β)x + αβ = 0

Sample Problems

Problem 1: Find the roots of the equation, 4x² + 5x + 1 = 0.

Solution:

Given equation is 4x² + 5x + 1 = 0

Now compare the given with the standard form ax² + bx + c = 0

So, we have a = 4, b = 5, c = 1

Now calculate the discriminant (D) = b² – 4ac = 5² – 4(4)(1) = 25 – 16 = 9 > 0

Since the discriminant is greater than zero, the given equation has two real roots.

x = [-b ± √(b² – 4ac)]/2a

= [-5 ± √9]/2(4)

= [-5 ± 3]/8 x = (-5 + 3)/8 & (-5 – 3)/8

= -2/8 & -8/8

= -1/4 & -1

Problem 2: Find the roots of the equation x² + 6x + 9 = 0.

Solution:

Given equation is x² +6x + 9 = 0.

Compare the given equation with the standard form ax² + bx + c = 0

So, we have a = 1, b = 6, c = 9

Now calculate the discriminant (D) = b² – 4ac = 6² – 4(1)(9) = 36 – 36 = 0

Since D = 0, the given equation has one real solution

x = [-b ± √(b² – 4ac)]/2a

= [-6 ± √0]/2(1)

= -6/2 

= -3

Problem 3: Find the roots of x² + 16 = 0.

Solution:

Given equation is x² + 16 = 0.

Now compare the given equation with the standard form ax² + bx + c = 0 

So, we have a = 1, b = 0, c = 16

Now calculate the discriminant (D) = b² – 4ac = 0² – 4(1)(16) = – 64 < 0

Since D < 0, the given equation has two complex roots

x = [-b ± √(b² – 4ac)]/2a

x = [0 ± √-64]/2(1)

x = ±8i/2, where ‘i’ is an imaginary number

x = 4i & -4i

Problem 4: Find the nature of roots of the quadratic equation 3x² + 6x + 4 = 0.

Solution:

Given equation is 3x² + 6x + 4 = 0

Now compare the equation with the standard form ax² + bx + c =0

So, we have a = 3, b = 4, c = 4

Now calculate the discriminant, 

D = b² – 4ac 

= 6² – 4(3)(4) 

= 36 – 48 = -12 < 0

Since discriminant (D) is less than zero, the equation has two imaginary solutions.

Problem 5: Find the sum and product of roots of the quadratic equation 3x² + 7x – 2 = 0.

Solution: 

Given, 3x² + 7x – 2 = 0

Now compare the given equation with the standard equation ax² + bx + c = 0

So, we have a = 3, b= 7 and c = -2

Let α and β be the roots of the given equation.

Sum of roots = -b/a = -7/3

Product of roots = c/a = -2/3

Problem 6: Find the quadratic equation if the sum of its roots = 3/4 and products of roots = 1.

Solution:

Given, sum of roots of the quadratic equation = α + β = 3/4

Product of roots of the quadratic equation = αβ = 1

Now the required equation having roots α and β is x² – (α + β)x + αβ = 0

x² – (3/4)x + 1 = 0 

4x² – 3x + 4 = 0 is the required quadratic equation.