Question

Check whether the following are the quadratic equations:
$\left( i \right){\left( {x + 1} \right)^2} = 2\left( {x - 3} \right)$
$\left( {ii} \right){x^2} - 2x = \left( { - 2} \right)\left( {3 - x} \right)$
$\left( {iii} \right)\left( {x - 2} \right)\left( {x + 1} \right) = \left( {x - 1} \right)\left( {x + 3} \right)$
$\left( {iv} \right)\left( {x - 3} \right)\left( {2x + 1} \right) = x\left( {x + 5} \right)$
$\left( v \right)\left( {2x - 1} \right)\left( {x - 3} \right) = \left( {x + 5} \right)\left( {x - 1} \right)$
$\left( {vi} \right){x^2} + 3x + 1 = {\left( {x - 2} \right)^2}$
$\left( {vii} \right){\left( {x + 2} \right)^3} = 2x\left( {{x^2} - 1} \right)$
$\left( {viii} \right){x^3} - 4{x^2} - x - 1 = {\left( {x - 2} \right)^3}$

Answer 1

Hint – In this particular question use the concept that the general representation of the quadratic equation is given as ($a{x^2} + bx + c = 0$), where a, b and c are some constants, it can be positive or negative, and (b) and (c) can be zero but (a) cannot be zero otherwise it cannot be a quadratic equation so use this concept to reach the solution of the question.

Complete step-by-step answer:
As we all know that the general quadratic equation is given as
$a{x^2} + bx + c = 0$, where a, b and c are some constants, it can be positive or negative, and (b) and (c) can be zero but (a) cannot be zero otherwise it cannot be a quadratic equation.
Now we have to check whether the given equations are quadratic equations or not.
$\left( i \right){\left( {x + 1} \right)^2} = 2\left( {x - 3} \right)$
Now open the square according to property ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ so we have,
$ \Rightarrow {x^2} + 2x + 1 = 2x - 6$
Now simplify this we have,
$ \Rightarrow {x^2} + 7 = 0$
So this represents the quadratic equation.
$\left( {ii} \right){x^2} - 2x = \left( { - 2} \right)\left( {3 - x} \right)$
Now simplify this equation we have,
$ \Rightarrow {x^2} - 2x = - 6 + 2x$
$ \Rightarrow {x^2} - 4x + 6 = 0$
So this also represents the quadratic equation.
$\left( {iii} \right)\left( {x - 2} \right)\left( {x + 1} \right) = \left( {x - 1} \right)\left( {x + 3} \right)$
Now simplify this equation we have,
$ \Rightarrow {x^2} + x - 2x - 2 = {x^2} + 3x - x - 3$
$ \Rightarrow 3x - 1 = 0$
As we see that the $\left( {{x^2}} \right)$ term is missing so this does not represents the quadratic equation.
 $\left( {iv} \right)\left( {x - 3} \right)\left( {2x + 1} \right) = x\left( {x + 5} \right)$
Now simplify this equation we have,
$ \Rightarrow 2{x^2} + x - 6x - 3 = {x^2} + 5x$
$ \Rightarrow {x^2} - 10x - 3 = 0$
So this also represents the quadratic equation.
$\left( v \right)\left( {2x - 1} \right)\left( {x - 3} \right) = \left( {x + 5} \right)\left( {x - 1} \right)$
Now simplify this equation we have,
$ \Rightarrow 2{x^2} + x - 6x - 3 = {x^2} - x + 5x - 5$
$ \Rightarrow {x^2} - 9x + 2 = 0$
So this also represents the quadratic equation.
$\left( {vi} \right){x^2} + 3x + 1 = {\left( {x - 2} \right)^2}$
Now open the square according to property ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$ so we have,
$ \Rightarrow {x^2} + 3x + 1 = {x^2} + 4 - 4x$
Now simplify this equation we have,
$ \Rightarrow 7x - 3 = 0$
As we see that the $\left( {{x^2}} \right)$ term is missing so this does not represent the quadratic equation.
$\left( {vii} \right){\left( {x + 2} \right)^3} = 2x\left( {{x^2} - 1} \right)$
Now open the cube according to the property ${\left( {a + b} \right)^3} = {a^3} + {b^3} + 3a{b^2} + 3{a^2}b$ so we have,
$ \Rightarrow {x^3} + 8 + 6{x^2} + 12x = 2{x^3} - 2x$
Now simplify this equation we have,
$ \Rightarrow - {x^3} + 6{x^2} + 14x + 8 = 0$
So as we see that $\left( {{x^3}} \right)$ term is present so it is a cubic equation not a quadratic equation.
$\left( {viii} \right){x^3} - 4{x^2} - x - 1 = {\left( {x - 2} \right)^3}$
Now open the cube according to the property ${\left( {a - b} \right)^3} = {a^3} - {b^3} + 3a{b^2} - 3{a^2}b$ so we have,
$ \Rightarrow {x^3} - 4{x^2} - x - 1 = {x^3} - 8 + 12x - 6{x^2}$
Now simplify this equation we have,
$ \Rightarrow 2{x^2} - 13x + 7 = 0$
So this also represents the quadratic equation.
So this is the required answer.

Note – Whenever we face such types of questions the key concept we have to remember is that always remember the general representation of the quadratic equation which is stated above then simplify these given equations one by one using ${\left( {a + b} \right)^3} = {a^3} + {b^3} + 3a{b^2} + 3{a^2}b$, ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$ as above we will get the required answer.