Question

Which of the following is not a quadratic equation?
a)\[{{(x-2)}^{2}}+1=2x-3\]
b)\[x(x+1)+8=\left( x-2 \right)\left( x-2 \right)\]
c)\[x(2x+3)={{x}^{2}}+1\]
d)\[{{(x-2)}^{3}}={{x}^{3}}-4\]

Answer 1

Hint: A quadratic equation is any equation that can be rearranged in standard form as \[a{{x}^{2}}+bx+c=0\] , where x represents an unknown, and a, b, and c represent known numbers, and a ≠ 0. If a = 0, then the equation is linear, not quadratic, as there is no \[{{x}^{2}}\] term. The numbers a, b, and c are the coefficients of the equation and may be distinguished by calling them, respectively, the quadratic coefficient, the linear coefficient and the constant or free term.

Complete step-by-step answer:
We will check every option for the non-quadratic equation. The equation which is not in the form of the standard 2-degree equation \[a{{x}^{2}}+bx+c=0\] where \[a\ne 0\] are not a quadratic equation.

Option a:

Given, \[{{\left( x-2 \right)}^{2}}+1=2x-3.\]

\[\Rightarrow {{x}^{2}}-4x+4+1=2x-3.\]

\[\Rightarrow {{x}^{2}}-4x+5=2x-3.\]

\[\Rightarrow {{x}^{2}}-6x+8=0.\]

Therefore, option a is a quadratic equation.

Option b:

Given, \[x(x+1)+8=\left( x-2 \right)\left( x-2 \right)\]

\[\Rightarrow {{x}^{2}}+x+8={{x}^{2}}-4x+4\]

\[\Rightarrow 5x+4=0\]

Here, the coefficient of \[{{x}^{2}}\] is zero. Therefore, the equation is a linear equation.
Hence, Option b is not a quadratic equation.

Option c:

Given, \[x(2x+3)={{x}^{2}}+1\]

\[\Rightarrow 2{{x}^{2}}+3x={{x}^{2}}+1\]

\[\Rightarrow {{x}^{2}}+3x-1=0\]

Therefore, option c is a quadratic equation.

Option d:

Given, \[{{(x-2)}^{3}}={{x}^{3}}-4\]

\[\Rightarrow {{x}^{3}}+6{{x}^{2}}-12x-8={{x}^{3}}-4\]

\[\Rightarrow 6{{x}^{2}}-12x-4=0\]

Therefore, option d is a quadratic equation.

Therefore, the correct option is option(b).

Note: Don’t get confused that in option (b) , the LHS has a quadratic coefficient which is not equal to zero, because the RHS also has a second-degree term with the same quadratic coefficient. The second-degree term will cancel out and will leave a linear equation. Hence, it will not be a quadratic equation.