Question

verify (2x+1)(3x+2)=6(x-1)(x-2) is a quadratic equation or not?explain​

Answer 1

EXPLANATION.

Equation.

⇒ (2x + 1)(3x + 2) = 6(x - 1)(x - 2).

As we know that,

Expand this equation, we get.

⇒ 2x(3x + 2) + 1(3x + 2) = 6[x(x - 2) - 1(x - 2)].

⇒ 6x² + 4x + 3x + 2 = 6[x² - 2x - (x - 2)].

⇒ 6x² + 7x + 2 = 6[x² - 2x - x + 2].

⇒ 6x² + 7x + 2 = 6[x² - 3x + 2].

⇒ 6x² + 7x + 2 = 6x² - 18x + 12.

⇒ 6x² + 7x + 2 - 6x² + 18x - 12 = 0.

⇒ 7x + 18x + 2 - 12 = 0.

⇒ 25x - 10 = 0.

As we can see that,

Quadratic equation is in the form of = ax² + bx + c = 0 (a ≠ 0).

This equation is not a quadratic equation.

                                                                                                                         

MORE INFORMATION.

Quadratic expression.

A polynomial of degree two of the form ax² + bx + c (a ≠ 0) is called a quadratic expression in x.

The quadratic equation.

ax² + bx + c = 0 (a ≠ 0) has two roots, given by.

⇒ α = - b + √D/2a.

⇒ β = - b - √D/2a.

D = Discriminant  or b² - 4ac.

Answer 2

$\implies\sf (2x + 1)(3x + 2) = 6(x - 1)(x - 2) \\ \\ \implies\sf2 {x}^{2} + 4x + 3x + 2 = 6( {x}^{2} - 2x - x + 2 ) \\ \\ \implies\sf6 {x}^{2} + 4x + 3x + 2 = 6( {x}^{2} - 3x + 2 ) \\ \\ \implies\sf \cancel{6 {x}^{2}} + 7x + 2 = \cancel{6 {x}^{2}} - 18x + 12 \\ \\ \implies\sf25x - 10 = 0$

As, this equation doesn't have the degree as 2 or the biggest power of the variable as 2. it is not a quadratic equation.

done :D