Question

Which of the following is not a quadratic equation?
A) \[2{(x + 1)^2} = 4{x^2} + 2x + 1\]
B) \[2x + {x^2} = 2{x^2} + 5\]
C) \[{\left( {\sqrt 2 \times \sqrt 3 x} \right)^2} + {x^2} = 3{x^2} + 5x\]
D) \[{\left( {{x^2} + 2x} \right)^2} = {x^5} + 3 + 4{x^3}\]

Answer 1

Hints: For a polynomial to be quadratic the necessary and sufficient condition is the highest power of the variable must be equal to 2. Try to solve all the given options to see if we can find any polynomial which has the highest power of a variable greater than or less than 2.

Complete Step by Step Solution:
For option A, \[2{(x + 1)^2} = 4{x^2} + 2x + 1\]
Using the formula \[{(a + b)^2} = {a^2} + 2ab + {b^2}\]
We can get,
\[\begin{array}{l}
2{(x + 1)^2} = 4{x^2} + 2x + 1\\
 \Rightarrow 2({x^2} + 2x + 1) = 4{x^2} + 2x + 1\\
 \Rightarrow 2{x^2} + 4x + 2 = 4{x^2} + 2x + 1\\
 \Rightarrow 2{x^2} - 2x - 1 = 0
\end{array}\]
Clearly the highest power of the variable x is 2. So it is not the correct option
For option B, \[2x + {x^2} = 2{x^2} + 5\]
\[\begin{array}{l}
2x + {x^2} = 2{x^2} + 5\\
 \Rightarrow {x^2} - 2x + 5 = 0
\end{array}\]
Again The highest power is 2. So it is not the Correct option
For option C, \[{\left( {\sqrt 2 \times \sqrt 3 x} \right)^2} + {x^2} = 3{x^2} + 5x\]
\[\begin{array}{l}
{\left( {\sqrt 2 \times \sqrt 3 x} \right)^2} + {x^2} = 3{x^2} + 5x\\
 \Rightarrow {\left( {\sqrt 6 x} \right)^2} + {x^2} = 3{x^2} + 5x\\
 \Rightarrow 6{x^2} + {x^2} = 3{x^2} + 5x\\
 \Rightarrow 7{x^2} = 3{x^2} + 5x\\
 \Rightarrow 4{x^2} - 5x = 0
\end{array}\]
And yet again we get the highest power as 2. Clearly it is not the solution we are looking for
So that’s make it obvious that option D is the correct solution, Still it can be elaborated as
\[\begin{array}{l}
{\left( {{x^2} + 2x} \right)^2} = {x^5} + 3 + 4{x^3}\\
 \Rightarrow {x^4} + 4{x^2} + 4{x^3} = {x^5} + 3 + 4{x^3}\\
 \Rightarrow {x^5} - {x^4} - 4{x^2} + 3 = 0
\end{array}\]
It is Clearly seen that the highest power of x is 5 here.
Therefore it is not a quadratic equation. Hence option D is correct

Note: Students always make mistakes while converting signs in solving equations so beware of them. On the other hand we could have reached the solution without doing all this in the last option highest power of 5 is clearly visible and by solving \[{\left( {{x^2} + 2x} \right)^2}\] we can only get a maximum of \[{x^{{2^2}}} = {x^4}\] therefore there was no chance that \[{x^5}\] would be cancelled out so there’s no chance of getting a quadratic equation.