Which of the following is not a quadratic equation:
$
(a)\,{\left( {x - 2} \right)^2} + 1 = 2x - 3 \\
(b){\text{ }}x\left( {x + 1} \right) + 8 = \left( {x + 2} \right)\left( {x - 2} \right) \\
(c){\text{ }}x\left( {2x + 3} \right) = {x^2} + 1 \\
(d){\text{ }}{\left( {x + 2} \right)^3} = {x^3} - 4 \\
$
Answer
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Hint – A quadratic equation is one which has the general form of $a{x^2} + bx + c = 0$ where $a \ne 0$. Simplify the given options step by step to check if it can be reduced to this general form or not.
Complete step-by-step answer:
As we know that in a quadratic the highest power of a variable is 2.
The general quadratic equation is $a{x^2} + bx + c = 0$
Now check options one by one we have,
$\left( i \right){\text{ }}{\left( {x - 2} \right)^2} + 1 = 2x - 3$
Now simplify this equation we have according to the formula${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$.
$ \Rightarrow {x^2} + 4 - 4x + 1 = 2x - 3$
$ \Rightarrow {x^2} - 6x + 8 = 0$
So this represents a quadratic equation.
Now check option (ii)
$\left( {ii} \right){\text{ }}x\left( {x + 1} \right) + 8 = \left( {x + 2} \right)\left( {x - 2} \right)$
Now simplify this equation we have,
$ \Rightarrow {x^2} + x + 8 = {x^2} - 4$
$ \Rightarrow x + 12 = 0$
So this does not represent the quadratic equation.
Now check option (iii)
$\left( {iii} \right){\text{ }}x\left( {2x + 3} \right) = {x^2} + 1$
Now simplify this equation we have,
$ \Rightarrow 2{x^2} + 3x = {x^2} + 1$
$ \Rightarrow {x^2} + 3x - 1 = 0$
So this also represents a quadratic equation.
Now check option (iv)
$\left( {iv} \right){\text{ }}{\left( {x + 2} \right)^3} = {x^3} - 4$
Now simplify this equation we have according to the formula${\left( {a + b} \right)^3} = {a^3} + {b^3} + 3a{b^2} + 3{a^2}b$.
$ \Rightarrow {x^3} + 8 + 6{x^2} + 12x = {x^3} - 4$
$ \Rightarrow 6{x^2} + 12x + 12 = 0$
Now divide by 6 we have,
$ \Rightarrow {x^2} + 2x + 2 = 0$
So this also represents a quadratic equation.
So from amongst options only option (B) does not represent the quadratic equation.
So option (B) is the required answer.
Note – In this question a basic algebraic identity of ${\left( {a + b} \right)^3}$is also used, we know that on solving this forms a cubic equation so directly we could have said that it won’t be quadratic . The trick to solve these types of problems is check for the highest power of x in the equation if it’s 2 that means the equation is quadratic else not.
Complete step-by-step answer:
As we know that in a quadratic the highest power of a variable is 2.
The general quadratic equation is $a{x^2} + bx + c = 0$
Now check options one by one we have,
$\left( i \right){\text{ }}{\left( {x - 2} \right)^2} + 1 = 2x - 3$
Now simplify this equation we have according to the formula${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$.
$ \Rightarrow {x^2} + 4 - 4x + 1 = 2x - 3$
$ \Rightarrow {x^2} - 6x + 8 = 0$
So this represents a quadratic equation.
Now check option (ii)
$\left( {ii} \right){\text{ }}x\left( {x + 1} \right) + 8 = \left( {x + 2} \right)\left( {x - 2} \right)$
Now simplify this equation we have,
$ \Rightarrow {x^2} + x + 8 = {x^2} - 4$
$ \Rightarrow x + 12 = 0$
So this does not represent the quadratic equation.
Now check option (iii)
$\left( {iii} \right){\text{ }}x\left( {2x + 3} \right) = {x^2} + 1$
Now simplify this equation we have,
$ \Rightarrow 2{x^2} + 3x = {x^2} + 1$
$ \Rightarrow {x^2} + 3x - 1 = 0$
So this also represents a quadratic equation.
Now check option (iv)
$\left( {iv} \right){\text{ }}{\left( {x + 2} \right)^3} = {x^3} - 4$
Now simplify this equation we have according to the formula${\left( {a + b} \right)^3} = {a^3} + {b^3} + 3a{b^2} + 3{a^2}b$.
$ \Rightarrow {x^3} + 8 + 6{x^2} + 12x = {x^3} - 4$
$ \Rightarrow 6{x^2} + 12x + 12 = 0$
Now divide by 6 we have,
$ \Rightarrow {x^2} + 2x + 2 = 0$
So this also represents a quadratic equation.
So from amongst options only option (B) does not represent the quadratic equation.
So option (B) is the required answer.
Note – In this question a basic algebraic identity of ${\left( {a + b} \right)^3}$is also used, we know that on solving this forms a cubic equation so directly we could have said that it won’t be quadratic . The trick to solve these types of problems is check for the highest power of x in the equation if it’s 2 that means the equation is quadratic else not.
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