Any equation of the form ${p(x) = 0}$, where ${p(x)}$ is a polynomial of degree
2, is a quadratic equation.
Standard form of a quadratic equation:Any quadratic equation of the form p(x) in descending order of their degrees is called standard form of a quadratic equation.
A quadratic equation in the variable ${x}$ is an equation of the form ${ax^2 + bx + c = 0}$, where ${a}$,${b}$, ${c}$ are real numbers, ${a ≠ 0}$.
For example, ${2x^2 + x – 300 = 0}$ is a quadratic equation.
Similarly, ${2x^2 – 3x + 1 = 0}$, ${4x – 3x^2 + 2 = 0 }$ and ${1 – x^2 + 300 = 0}$ are also quadratic equations.
Example :Represent the following situations mathematically:
(i)John and Jivanti together have 45 marbles. Both of them lost 5 marbles each,
and the product of the number of marbles they now have is 124.
We would like to find out how many marbles they had to start with.
Solution:
(i) Let the number of marbles John had be ${x}$.
Since, John and Jivanti together have 45 marbles.
Therefore,the number of marbles Jivanti had = ${45-x}$ .
The number of marbles left with John, when he lost 5 marbles = ${x-5}$
The number of marbles left with Jivanti, when she lost 5 marbles = ${45 – x – 5}$ = ${40-x}$
The product of the number of marbles they now have is 124.
Therefore,
${(x-5)}$ x ${(40-x)}$=124
${40x – x^2 – 200 + 5x}$=124
${x^2 – 45x + 324 = 0}$
Therefore, the number of marbles John had, satisfies the quadratic equation
${x^2 – 45x + 324 = 0}$
which is the required representation of the problem mathematically.
Example:
(ii)A cottage industry produces a certain number of toys in a day.
The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day,
the total cost of production was ₹ 750. Find out the number of toys produced on that day.
Solution:
Let the number of toys produced on that day be ${x}$.
Therefore, the cost of production (in rupees) of each toy that day = ${55 – x}$
So, the total cost of production (in rupees) that day = ${x (55 – x)}$
On a particular day, the total cost of production was ₹ 750
Therefore,
${x (55 – x)}$ = 750
i.e., ${55x – x2}$ = 750
i.e., ${– x^2 + 55x – 750 = 0}$
i.e., ${x^2 – 55x + 750 = 0}$
Therefore, the number of toys produced that day satisfies the quadratic equation
${x^2 – 55x + 750 = 0}$
which is the required representation of the problem mathematically.
Example: Check whether the following are quadratic equations:
(i) ${(x – 2)^2 + 1 = 2x – 3 }$
(ii) ${x(x + 1) + 8 = (x + 2) (x – 2)}$
(iii) ${x (2x + 3) = x^2 + 1}$
(iv) ${(x + 2)^3 = x^3 – 4}$
Solution :
(i) LHS = ${(x – 2)^2 + 1}$
= ${x^2 – 4x + 4 + 1}$
= ${x^2 – 4x + 5}$
Therefore, ${(x – 2)^2 + 1 = 2x – 3}$ can be rewritten as
${x^2 – 4x + 5 = 2x – 3}$
i.e., ${x^2 – 6x + 8 = 0}$
It is of the form ${ax^2 + bx + c = 0}$.
Therefore, the given equation is a quadratic equation.
(ii) Since ${x(x + 1) + 8 = x^2 + x + 8}$ and ${(x + 2)(x – 2) = x^2 – 4}$
Therefore, ${x^2 + x + 8 = x^2 – 4}$
i.e., ${x + 12 = 0}$
It is not of the form ${ax^2 + bx + c = 0.}$
Therefore, the given equation is not a quadratic equation.
(iii) Here, LHS = ${x (2x + 3) = 2x^2 + 3x}$
So, ${x (2x + 3) = x^2 + 1}$ can be rewritten as
${2x^2 + 3x = x^2 + 1}$
Therefore, we get ${x^2 + 3x – 1 = 0}$
It is of the form ${ax^2 + bx + c = 0}$.
So, the given equation is a quadratic equation.
(iv) Here,LHS = ${(x + 2)^3}$ = ${x^3 + 6x^2 + 12x + 8}$
Therefore, ${(x + 2)^3}$ = ${x^3 – 4}$ can be rewritten as
${x^3 + 6x^2 + 12x + 8 = x^3 – 4}$
i.e., ${6x^2 + 12x + 12 = 0}$ or, ${x^2 + 2x + 2 = 0}$
It is of the form ${ax^2 + bx + c = 0}$.
So, the given equation is a quadratic equation.
Remark : Be careful! In (ii) above, the given equation appears to be a quadratic equation,
but it is not a quadratic equation.
In (iv) above, the given equation appears to be a cubic equation (an equation of degree 3)
and not a quadratic equation. But it turns out to be a quadratic equation.
As you can see, often we need to simplify the given equation before deciding whether it is quadratic or not.
