# 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 = -$$\frac{1}{4}$$

Algebra/Linear Algebra

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