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
General Properties of Quadratic Equation
Methods of Solving Quadratic Equations
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
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