Examine whether the following equation is quadratic or not: $(2x + 1)(3x + 2) = 6(x - 1)(x - 2)$
(B) To determine if the equation is quadratic,we simplify both sides:
Left Hand Side $(LHS)$: $(2x + 1)(3x + 2) = 6x^2 + 4x + 3x + 2 = 6x^2 + 7x + 2$
Right Hand Side $(RHS)$: $6(x - 1)(x - 2) = 6(x^2 - 2x - x + 2) = 6(x^2 - 3x + 2) = 6x^2 - 18x + 12$
Equating $LHS$ and $RHS$: $6x^2 + 7x + 2 = 6x^2 - 18x + 12$
Subtracting $6x^2$ from both sides: $7x + 2 = -18x + 12$
Rearranging the terms: $7x + 18x + 2 - 12 = 0$
$25x - 10 = 0$
Since the highest power of the variable $x$ is $1$ and not $2$,the equation is a linear equation,not a quadratic equation.
Left Hand Side $(LHS)$: $(2x + 1)(3x + 2) = 6x^2 + 4x + 3x + 2 = 6x^2 + 7x + 2$
Right Hand Side $(RHS)$: $6(x - 1)(x - 2) = 6(x^2 - 2x - x + 2) = 6(x^2 - 3x + 2) = 6x^2 - 18x + 12$
Equating $LHS$ and $RHS$: $6x^2 + 7x + 2 = 6x^2 - 18x + 12$
Subtracting $6x^2$ from both sides: $7x + 2 = -18x + 12$
Rearranging the terms: $7x + 18x + 2 - 12 = 0$
$25x - 10 = 0$
Since the highest power of the variable $x$ is $1$ and not $2$,the equation is a linear equation,not a quadratic equation.
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