Solution of a Quadratic Equation by Factorisation
A real number ${α}$ is called a root of the quadratic equation
${ax^2 + bx + c = 0, a ≠ 0}$ if ${a α^2 + bα + c = 0}$.
We also say that ${x = α}$ is a solution of the quadratic equation,
or that ${α}$ satisfies the quadratic equation.
Note that the zeroes of the
quadratic polynomial ${ax^2 + bx + c}$ and the roots of the quadratic equation ${ax^2 + bx + c = 0}$
are the same.
Example : Find the roots of the equation ${2x^2 – 5x + 3 = 0}$, by factorisation.
Solution : Split the middle term ${– 5x}$ as ${–2x}$,${ –3x}$ [because ${(–2x) × (–3x) = 6x^2 = (2x^2) × 3}$]. So, ${2x^2 – 5x + 3}$ = ${2x^2 – 2x – 3x + 3}$ = ${2x (x – 1) –3(x – 1)}$ = ${(2x – 3)(x – 1)}$ Now, ${2x^2 – 5x + 3 = 0}$ can be rewritten as ${(2x – 3)(x – 1) = 0}$. So, the values of ${x}$ for which ${2x^2 – 5x + 3 = 0}$ are the same for which ${(2x – 3)(x – 1) = 0}$, i.e., either ${2x – 3 = 0}$ or ${x – 1 = 0}$. i.e ${x=3/2}$ or ${x=1}$ Answer: So,${x=3/2}$ or ${x=1}$ are the roots of the equation.
Example : Find the roots of the quadratic equation ${6x^2 – x – 2 = 0}$.Solution : ${6x^2 – x – 2 = 6x^2 + 3x – 4x – 2}$ =${3x (2x + 1) – 2 (2x + 1)}$ =${(3x – 2)(2x + 1)}$ The roots of ${6x^2 – x – 2 = 0}$ are the values of ${x}$ for which ${(3x – 2)(2x + 1) = 0}$ Therefore, ${3x – 2 = 0}$ or ${2x + 1 = 0,}$ i.e ${x=2/3}$ or ${x=-1/2}$ Therefore, the roots of ${6x^2 – x – 2 = 0}$ are ${2/3}$ or ${-1/2}$Example : Find the roots of the quadratic equation ${3x^2 − 2√6x + 2 = 0}$ .Solution : ${3x^2 − 2√6x + 2 }$ = ${3x^2 − √6x - √6x + 2 }$ = ${√3x(√3x-√2) - √2(√3x-√2)}$ = ${(√3x-√2)(√3x-√2)}$ Now ${(√3x-√2)=0}$, therefore ${x=√(2/3)}$ Since the roots are repeated twice, Roots of ${3x^2 − 2√6x + 2 = 0}$ are ${x=√(2/3)}$ , ${x=√(2/3)}$