Math, asked by manognarachakonda32, 1 month ago

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

Answers

Answered by amansharma264
10

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.


Anonymous: Awesome bhai ! :)
amansharma264: Thanku
Answered by AestheticSky
25

  \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

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