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 = \(\frac{1}{4}\)
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