[Maths Class Notes] on Quadratic Formula Pdf for Exam

An equation of the form ax2 + bx + c = 0, where a, b, c are real numbers and a0 is called a quadratic equation. The value of unknown variable x, which satisfies the given quadratic equation is called the roots of quadratic equation. For example, if is a root of quadratic equation ax2 + bx + c = 0, then a2 + b + c = 0. And the process of finding roots is known as solving a quadratic equation. There are three methods to solve a quadratic equation, which are as follows:

1. Solving a quadratic equation by factorisation.

2. Solving a quadratic equation by completing the square.

3. Solving a quadratic equation using quadratic formula.

In this article, we will learn about quadratic formula, its derivation and how to solve a quadratic equation using quadratic formula. 

What is Quadratic Formula?

Quadratic formula is a formula that helps us to find the roots of a quadratic equation very easily by replacing the other methods of finding the roots like, factorisation method, completing the square method. The quadratic formula to find the roots of quadratic equation ax2 + bx + c = 0, where a ≠ 0 is given by:

[x = frac{-b pm sqrt{b^{2} – 4ac}}{2a}]

The plus (+) and minus (-) sign represents that quadratic formula will give two roots, one root corresponding to the plus (+) sign and another root corresponding to the minus (-) sign.

i.e.,  [x_{1} = frac{-b – sqrt{b^{2} – 4ac}}{2a}] and [x_{2} = frac{-b + sqrt{b^{2} – 4ac}}{2a}]. Both roots are evaluated by substituting the corresponding values of coefficients a, b and c from the quadratic equation ax2 + bx + c = 0.

Derivation of Quadratic Formula

Consider the quadratic equation ax2 + bx + c = 0, where a, b, c are real numbers and a0. Then,

ax2 + bx + c = 0

⇒ ax2 + bx = –c

On dividing both sides by a, we get:

⇒ [x^{2} + frac{b}{a}x = frac{-c}{a}]

Now adding [(frac{b}{2a})^{2}] on both sides, we get:

⇒ [x^{2} + frac{b}{a}x + (frac{b}{2a})^{2} = frac{-c}{a} + (frac{b}{2a})^{2}]

⇒ [(x  + frac{b}{2a})^{2} = (frac{-c}{a} + frac{b^{2}}{4a^{2}})]

⇒ [(x + frac{b}{2a})^{2} = frac{(b^{2} – 4ac)}{4a^{2}}]

⇒ [(x – frac{b}{2a}) = frac{pm sqrt{b^{2} – 4ac}}{2a}], when (b2 – 4ac) 0

⇒ [x = – frac{b}{2a} pm frac{sqrt{b^{2} – 4ac}}{2a}]

⇒ [x = frac{-b pm sqrt{b^{2} – 4ac}}{2a}]

Therefore, the roots of quadratic equation ax2 + bx + c = 0 is  [x_{1} = frac{-b + sqrt{b^{2} – 4ac}}{2a}] and [x_{2} = frac{-b – sqrt{b^{2} – 4ac}}{2a}].

Discriminant of Quadratic Equation

The expression D = (b2 – 4ac) is called the discriminant of quadratic equation ax2 + bx + c = 0. 

The roots of ax2 + bx + c = 0 are real only when D 0 i.e., b2 – 4ac 0 and the roots are given by [x_{1} = frac{-b + sqrt{D}}{2a}] and [x_{2} = frac{-b – sqrt{D}}{2a}].

How to Solve Quadratic Equations using Quadratic Formula?

To learn how to solve quadratic equations using quadratic formula, let us consider some examples and solve them using quadratic formula.

Example 1: Find the roots of quadratic equation 15x2 – x – 28 = 0 using quadratic formula.

Solution:

The given quadratic equation is 15x2 – x – 28 = 0. Comparing it with ax2 + bx + c = 0, we get a = 15, b = -1 and c = -28.

So, D = b2 – 4ac = (-1)2 – 4 × 15 × (-28) = 1681. As D = 1681 > 0, The given quadratic equation has real roots.

Now substituting the corresponding values of a, b and c in quadratic formula: [x = frac{-b pm sqrt{b^{2} – 4ac}}{2a}], we get,

[x = frac{-(-1) pm sqrt{(-1)^{2} – 4 times 15 times (-28)}}{2 times 15}]

⇒ [x = frac{1 pm sqrt{1681}}{30}]

For plus (+) sign, 

The root is [x_{1} = frac{1 + sqrt{1681}}{30} = frac{1 + 41}{30} = frac{42}{30} = frac{7}{5}].

and, for minus (-) sign, 

the root is [x_{2} = frac{1 – sqrt{1681}}{30} = frac{1 – 41}{30} = frac{-40}{30} = frac{-4}{3}]

Hence, the required roots of quadratic equation 15x2 – x – 28 = 0 are 7/5 and -4/3 .

Example 2: Find the roots of quadratic equation x2 + 6x + 6 = 0 using quadratic formula.

Solution:

The given quadratic equation is x2 + 6x + 6 = 0. Comparing it with ax2 + bx + c = 0, we get a = 1, b = 6 and c = 6.

So, D = b2 – 4ac = (6)2 – 4 × 1 × 6 = 12. As D = 12 > 0, The given quadratic equation has real roots.

Now substituting the corresponding values of a, b and c in quadratic formula: [x = frac{-b pm sqrt{b^{2} – 4ac}}{2a}], we get,

[x = frac{-(-6) pm sqrt{(6)^{2} – 4 times 1 times 6}}{2 times 1}]

⇒ [x = frac{-6 pm sqrt{12}}{2}]

For plus (+) sign, 

the root is [x_{1} = frac{-6 + sqrt{12}}{2} = frac{-6 + 2sqrt{3}}{2} = (-3 + sqrt{3})]

and, for minus (-) sign, 

the root is the root is [x_{2} = frac{-6 – sqrt{12}}{2} = frac{-6 – 2sqrt{3}}{2} = (-3 – sqrt{3})] 

Hence, the required roots of quadratic equation x2 + 6x + 6 = 0 are [((-3 + sqrt{3})] and  [(-3 – sqrt{3})].