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Roots of a Quadratic Equation

We will learn how to find the Roots of a quadratic equation.

Every quadratic equation gives two values of the unknown variable and these values are called roots of the equation.

Let ax2 + bx + c = 0 be a quadratic equation. If aα2 + bα + c = 0 then α is called a root of the quadratic equation ax2 + bx + c = 0.

Thus,

α is a root of ax2 + bx + c = 0 if and only if aα2 + bα + c = 0

If aα2 + bα + c = 0 then we say x = α satisfies the equation ax2 + bx + c = 0 and x = α is a solution.

Thus, every solution is root.

A quadratic equation has two roots which may be unequal real numbers or equal real numbers, or numbers which are not real.

If a quadratic equation has two real equal roots α, we say the equation has only one real solution.


Example: Let 3x2 + x - 2 = 0 be a quadratic equation. Clearly,

3 ∙ (-1)2 + (-1) - 2 = 0

So, x = -1 is a root of the quadratic equation 3x2 + x - 2 = 0.

Similarly, x = 2/3 is another root of the equation.

But x = 2 is not a root of 3x2 + x - 2 = 0 because 3 ∙ 22 + 2 - 2 ≠ 0.


Solved examples to find the roots of a quadratic equation:

1. Without solving the quadratic equation 3x2 - 2x - 1 = 0, find whether x = 1 is a solution (root) of this equation or not.

Solution:

Substituting x = 1 in the given equation 3x2 - 2x - 1 = 0, we get

3(1)2 - 2 (1) - 1 = 0

⟹ 3 - 2 - 1 = 0

⟹ 3 - 3 = 0; which is true.

Therefore, x = 1 is a solution of the given equation 3x2 - 2x - 1 = 0


2. Without solving the quadratic equation x2 - x + 1 = 0, find whether x = -1 is a root of this equation or not.

Solution:

Substituting x = -1 in the given equation x2 - x + 1 = 0, we get

(-1)2 - (-1) + 1 = 0

⟹ 1 + 1 + 1 = 0

⟹ 3 = 0; which is not true.

Therefore, x = -1 is not a solution of the given equation x2 - x + 1 = 0.

 

3. If one root of the quadratic equation 2x2 + ax - 6 = 0 is 2, find the value of a. Also, find the other root.

Solution:

Since, x = 2 is a root of the gives equation 2x2 + ax - 6 = 0

⟹ 2(2)2 + a × 2 - 6 = 0

⟹ 8 + 2a - 6 = 0

⟹ 2a + 2 = 0

⟹ 2a = -2

⟹ a = 22

⟹ a = -1

Therefore, the value of a = -1

Substituting a = -1, we get:

2x2 + (-1)x - 6 = 0

⟹ 2x2 - x - 6 = 0

⟹ 2x2 - 4x + 3x - 6 = 0

⟹ 2x(x - 2) + 3(x - 2) = 0

⟹ (x - 2)(2x + 3) = 0

⟹ x - 2 = 0 or 2x + 3 = 0

i.e., x = 2 or x = -32

Therefore, the other root is -32.


4. Find the value of k for which x = 2 is a root (solution) of equation kx2 + 2x - 3 = 0.

Solution:

Substituting x = 2 in the given equation kx2 + 2x - 3 = 0; we get:

K(2)2 + 2 × 2 - 3 = 0

⟹ 4k + 4 - 3 = 0

⟹ 4k + 1 =

⟹ 4k = -1

⟹ k = -14

Therefore, the value of k = -14

Quadratic Equation

Introduction to Quadratic Equation

Formation of Quadratic Equation in One Variable

Solving Quadratic Equations

General Properties of Quadratic Equation

Methods of Solving Quadratic Equations

Roots of a Quadratic Equation

Examine the Roots of a Quadratic Equation

Problems on Quadratic Equations

Quadratic Equations by Factoring

Word Problems Using Quadratic Formula

Examples on Quadratic Equations 

Word Problems on Quadratic Equations by Factoring

Worksheet on Formation of Quadratic Equation in One Variable

Worksheet on Quadratic Formula

Worksheet on Nature of the Roots of a Quadratic Equation

Worksheet on Word Problems on Quadratic Equations by Factoring









9th Grade Math

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