4.1 Introduction

The quadratic polynomial of the form $$ax^{2} + bx + c$$,  c, a ≠ 0. When we equate this polynomial to zero, we get a quadratic equation. Quadratic equations come up when we deal with many real-life situations.

suppose a charity trust decides to build a prayer hall having a carpet area of 300 square metres with its length one metre more than twice its breadth.

Suppose the breadth of the hall is x metres. Then, its length should be (2x + 1) metres We can depict this information pictorially 

Now, area of the hall = (2x + 1). $$x m^{2} = (2x^{2} + x) m^{2}$$

So, $$2x^{2} + x = 300$$           (Given)

Therefore,   $$2x^{2}+x-300=0$$

So, the breadth of the hall should satisfy the equation $$2x^{2} + x – 300 = 0$$ which is a quadratic equation.

4.2 Quadratic Equations

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, $$2x^{2} – 3 x + 1 = 0$$, $$4x – 3x^{2} + 2 = 0$$ and $$1 – x^{2} + 300 = 0$$ are also quadratic equations.

In fact, any equation of the form p (x) = 0, where p (x) is a polynomial of degree 2, is a quadratic equation. But when we write the terms of p (x) in descending order of their degrees, then we get the standard form of the equation. That is, $$ax^{2} + bx + c = 0$$, a ≠ 0 is called the standard form of a quadratic equation.

Example 1 : 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.

(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. We would like to find out the number of toys produced on that day.

Solution :

(i) Let the number of marbles John had be x.

Then the number of marbles Jivanti had = 45 – x (Why?).

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

Therefore, their product = (x – 5) (40 – x)

=$$40x – x^{2} – 200 + 5x$$

= $$– x^{2} + 45x – 200$$ So, $$– x^{2} + 45x – 200 = 124$$ (Given that product = 124)

i.e.,$$ – x^{2} + 45x – 324 = 0$$

i.e., $$x^{2} – 45x + 324 = 0$$

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.

(ii)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)

Therefore, x (55 – x) = 750

i.e., 55x – x 2 = 750

i.e., – x 2 + 55x – 750 = 0 i.e., x2 – 55x + 750 = 0

Therefore, the number of toys produced that day satisfies the quadratic equation x2 – 55x + 750 = 0

which is the required representation of the problem mathematically.

Example 2 : 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) x (x + 1) + 8 = (x + 2) (x – 2)

Sol.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)$$x(2x+3)=x^{2}+1$$

Sol.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 $$ax2 + bx + c = 0$$.

So, the given equation is a quadratic equation.

iv)$$(x+2)^{3}=x^{3}-4$$

Sol.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.