Question

Which of the following is a quadratic equation? Select the correct option:
(A). \[{{x}^{1/2}}+2x+3=0\]
(B). \[(x-1)(x+4)={{x}^{2}}+1\]
(C). \[{{x}^{2}}-3x+5=0\]
(D). \[(2x+1)(3x-4)=6{{x}^{2}}+3\]

Answer 1

Hint: It is very important that we simplify the given equation to check if it is the quadratic equation or not. The quadratic equation will be of the form \[a{{x}^{2}}+bx+c=0\], where \[a\ne 0\].

Complete step-by-step solution -
In the question we have selected the quadratic equation from the given options.
Now, we know that the equation is in general form can be written as \[a{{x}^{2}}+bx+c=0\], where \[a\ne 0\]. Here, b and c can be zero or any non-zero value.
So we will take the first equation in option A, as follows:
\[\Rightarrow {{x}^{1/2}}+2x+3=0\]
So this equation is not in the form of \[a{{x}^{2}}+bx+c=0\]. So, it is not a quadratic equation.
Next we will take the equation \[(x-1)(x+4)={{x}^{2}}+1\] in option B.
Expanding the left hand side we will get:
\[\begin{align}
  & \Rightarrow (x-1)(x+4)={{x}^{2}}+1 \\
 & \Rightarrow xx+x\cdot \;4+\left( -1 \right)x+\left( -1 \right)\cdot \;4={{x}^{2}}+1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\because (a+b)(c+d)=ac+ad+bc+bd \\
 & \Rightarrow {{x}^{2}}+3x-4={{x}^{2}}+1 \\
 & \Rightarrow {{x}^{2}}+3x-{{x}^{2}}={{x}^{2}}+5-{{x}^{2}} \\
 & \Rightarrow \text{3x=5} \\
\end{align}\]
Now, \[\text{3x=5}\] is a linear equation and is not the quadratic equation as it is not in the form of \[a{{x}^{2}}+bx+c=0\].
So, \[(x-1)(x+4)={{x}^{2}}+1\]is not a quadratic equation but a linear equation.
Next, we will take the equation \[{{x}^{2}}-3x+5=0\] in option C and this can be clearly seen that it is of the form \[a{{x}^{2}}+bx+c=0\] where \[a=1\]
So, \[{{x}^{2}}-3x+5=0\]is a quadratic equation.
Next, we will take the equation \[(2x+1)(3x-4)=6{{x}^{2}}+3\] in option D.
Expanding the left hand side we will get:
\[\begin{align}
  & \Rightarrow (2x+1)(3x-4)=6{{x}^{2}}+3 \\
 & \Rightarrow 2\cdot \;3xx-2\cdot \;4x+1\cdot \;3x-1\cdot \;4=6{{x}^{2}}+3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\because (a+b)(c+d)=ac+ad+bc+bd \\
 & \Rightarrow 6{{x}^{2}}-5x-4=6{{x}^{2}}+3 \\
 & \Rightarrow 6{{x}^{2}}-5x-6{{x}^{2}}=6{{x}^{2}}+7-6{{x}^{2}} \\
 & \Rightarrow \text{-5x=7} \\
\end{align}\]
Now, \[\text{-5x=7}\] is a linear equation and is not the quadratic equation as it is not in the form of \[a{{x}^{2}}+bx+c=0\].
So, \[(2x+1)(3x-4)=6{{x}^{2}}+3\]is not a quadratic equation but a linear equation.
Hence we have only one quadratic equation and that is option C) \[{{x}^{2}}-3x+5=0\].

Note: When simplifying the equation, make sure there is no calculation error and use the correct expansion formula. The expansion formula for the expression of the form \[(a+b)(c+d)=ac+ad+bc+bd\] is used to simplify. While solving the quadratic equation, we need to bring the equation in the form \[a{{x}^{2}}+bx+c=0\].