# 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 ax$$^{2}$$ + bx + c = 0 be a quadratic equation. If aα$$^{2}$$ + bα + c = 0 then α is called a root of the quadratic equation ax$$^{2}$$ + bx + c = 0.

Thus,

α is a root of ax$$^{2}$$ + bx + c = 0 if and only if aα$$^{2}$$ + bα + c = 0

If aα$$^{2}$$ + bα + c = 0 then we say x = α satisfies the equation ax$$^{2}$$ + 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 3x$$^{2}$$ + x - 2 = 0 be a quadratic equation. Clearly,

3 ∙ (-1)$$^{2}$$ + (-1) - 2 = 0

So, x = -1 is a root of the quadratic equation 3x$$^{2}$$ + x - 2 = 0.

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

But x = 2 is not a root of 3x$$^{2}$$ + x - 2 = 0 because 3 ∙ 2$$^{2}$$ + 2 - 2 ≠ 0.

Solved examples to find the roots of a quadratic equation:

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

Solution:

Substituting x = 1 in the given equation 3x$$^{2}$$ - 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 3x$$^{2}$$ - 2x - 1 = 0

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

Solution:

Substituting x = -1 in the given equation x$$^{2}$$ - 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 x$$^{2}$$ - x + 1 = 0.

3. If one root of the quadratic equation 2x$$^{2}$$ + 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 2x$$^{2}$$ + ax - 6 = 0

⟹ 2(2)$$^{2}$$ + a × 2 - 6 = 0

⟹ 8 + 2a - 6 = 0

⟹ 2a + 2 = 0

⟹ 2a = -2

⟹ a = $$\frac{-2}{2}$$

⟹ a = -1

Therefore, the value of a = -1

Substituting a = -1, we get:

2x$$^{2}$$ + (-1)x - 6 = 0

⟹ 2x$$^{2}$$ - x - 6 = 0

⟹ 2x$$^{2}$$ - 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 = -$$\frac{3}{2}$$

Therefore, the other root is -$$\frac{3}{2}$$.

4. Find the value of k for which x = 2 is a root (solution) of equation kx$$^{2}$$ + 2x - 3 = 0.

Solution:

Substituting x = 2 in the given equation kx$$^{2}$$ + 2x - 3 = 0; we get:

K(2)$$^{2}$$ + 2 × 2 - 3 = 0

⟹ 4k + 4 - 3 = 0

⟹ 4k + 1 =

⟹ 4k = -1

⟹ k = -$$\frac{1}{4}$$

Therefore, the value of k = -$$\frac{1}{4}$$

Formation of Quadratic Equation in One Variable

Word Problems on Quadratic Equations by Factoring

Worksheet on Formation of Quadratic Equation in One Variable

Worksheet on Nature of the Roots of a Quadratic Equation

Worksheet on Word Problems on Quadratic Equations by Factoring