Question

quadratic equation using factorisation method

Answer 1

Hey there!

Factorizing the equation to find the solutions by using quadratic formula:

$\bf{(3x - 2) (x + 4) = (2x - 3) (3x - 8)}$

$\bf{3xx + 3x \times 4x - 2x - 2 \times 4 = 2 \times 3xx - 2 \times 8x - 3 \times 3x + 3 \times 8}$

$\bf{\underline{3x^2 + 10x - 8 = 6x^2 - 25x + 24}}$

Now, simplify by subtracting by "24", adding "25x", Subtracting "6x^2", then simplifying to get this quadratic equation:

$\bf{\underline{- 3x^2 + 35x - 32 = 0}}$

Solving it by applying quadratic formula for equation "3x^2 + 35x - 32 = 0"

For a required quadratic equation of $\bf{ax^2 + bx + c = 0}$ the solutions for which can be represented by the quadratic formula of :

$\boxed{\bf{x_{1, \: 2} = \dfrac{- b +- \sqrt{b^2 - 4ac}}{2a}}}$

Here, a = - 3,  b = 35,  c = - 32.

Solving for positive and negative values respectively:

$\bf{x_{1} = \dfrac{- 35 - \sqrt{35^2 - 4(3)(32)}}{- 2(3)}}$

$\bf{x_{1} = \dfrac{- 35 - \sqrt{1225 - 384}}{- 2 \times 3}}$

$\bf{x_{1} = \dfrac{- 35 - \sqrt{841}}{- 2 \times 3}}$

$\bf{x_{1} = \dfrac{- 35 - \sqrt{841}}{- 6}}$

$\bf{x_{1} = \dfrac{35 + 29}{6}}$

$\bf{x_{1} = \dfrac{64}{6}}$

$\bf{\therefore \quad x_{1} = \dfrac{32}{3}}$

For the second solution in Positive form of equation:

$\bf{x_{2} = \dfrac{- 35 + \sqrt{35^2 - 4(3)(32)}}{- 2(3)}}$

$\bf{x_{2} = \dfrac{- 35 + \sqrt{1225 - 384}}{- 2 \times 3}}$

$\bf{x_{2} = \dfrac{- 35 + \sqrt{841}}{- 2 \times 3}}$

$\bf{x_{2} = \dfrac{- 35 + \sqrt{841}}{- 6}}$

$\bf{x_{2} = \dfrac{35 - 29}{6}}$

$\bf{x_{2} = \dfrac{6}{6}}$

$\bf{\therefore \quad x_{2} = 1}$

Therefore the final solutions for this quadratic equations are:

$\boxed{\underline{\bf{x_{1} = \dfrac{32}{3}}}}$

$\boxed{\underline{\bf{x_{2} = 1}}}$

Hope this helps you and solves the doubts for getting the equation solved by quadratic method!