Question

$6x - \frac{2}{x } = 1$
solve the quadratic equations by factorisation method​

Answer 1

EXPLANATION.

Quadratic equation.

⇒ 6x - 2/x = 1.

As we know that,

Factorizes the equation, we get.

⇒ 6x² - 2 = x.

⇒ 6x² - x - 2 = 0.

Factorizes the equation into middle term splits, we get.

⇒ 6x² - 4x + 3x - 2 = 0.

⇒ 2x(3x - 2) + 1(3x - 2) = 0.

⇒ (2x + 1)(3x - 2) = 0.

⇒ x = -1/2  and  x = 2/3.

                                                                                                                     

MORE INFORMATION.

Nature of the factors of the quadratic expression.

(1) = Real and different, if b² - 4ac > 0.

(2) = Rational and different, if b² - 4ac is a perfect square.

(3) = Real and equal, if b² - 4ac = 0.

(4) = If D < 0 Roots are imaginary and unequal or complex conjugate.

Answer 2

${\pmb{\underline{\sf{ Required \ Solution... }}}}$

We have to solve using Quadratic Equation as:-

$\colon\implies{\sf{ 6x - \dfrac{2}{x} = 1 }} \\ \\ \\ \colon\implies{\sf{ \dfrac{6x^2 - 2}{x} = 1 }} \\ \\ \\ \colon\implies{\sf{ 6x^2 - 2 = x }} \\ \\ \\ \colon\implies{\sf{ 6x^2 - 2 - x = 0 }} \\ \\ \\ \colon\implies{\sf{ 6x^2 - x - 2 = 0 \ \ \ \ \ (Arranging) }} \\ \\ \\ \colon\implies{\sf{ 6x^2 - (4-3)x-2 = 0 }} \\ \\ \\ \colon\implies{\sf{ 6x^2 - 4x + 3x - 2 = 0 }} \\ \\ \\ \colon\implies{\sf{ 2x(3x-2) +1(3x-2) = 0 }} \\ \\ \\ \colon\implies{\sf{ (3x-2)(2x+1) = 0 }} \\ \\ \\ \colon\implies{\underline{\boxed{\sf\red{ x = \dfrac{2}{3} \ and \ - \dfrac{1}{2} }}}}$

$\\ \\ {\pmb{\underline{\sf{ More \ to \ Know... }}}}$

${\pmb{\underline{\sf{ As \ Explained \ Example... }}}}$

$\bigstar$ The roots of the quadratic equation: x = (-b ± √D)/2a

• where D = b² – 4ac

$\bigstar$ Nature of roots:

• D > 0, roots are real and distinct (unequal)
• D = 0, roots are real and equal (coincident)
• D < 0, roots are imaginary and unequal

$\bigstar$ The roots (α + iβ), (α – iβ) are the conjugate pair of each other.

$\bigstar$ Sum and Product of roots: If α and β are the roots of a quadratic equation,

then

• S = α+β= -b/a = coefficient of x/coefficient of x²
• P = αβ = c/a = constant term/coefficient of x²

$\bigstar$ Quadratic equation in the form of roots: x² – (α+β)x + (αβ) = 0