1.   Check whether the following are quadratic equations :

(i)    ${(x + 1)^2 = 2(x – 3)}$ 
 Solution   
$x^2+2x+1$ = $2x-6$

$x^2+7=0$    

Since it is a polynomial of degree 2, so it is a quadratic polynomial.       

(ii)  ${x^2 – 2x = (–2) (3 –x)}$
 Solution  
${x^2-2x=-6+2x}$  
${x^2-4x+6 = 0}$  
Since it is a polynomial of degree 2, so it is a quadratic polynomial.

(iii) ${(x – 2)(x + 1) = (x – 1)(x + 3)}$ 

 Solution  
${x^2+x-2x-2}$ = $x^2+3x-x-3$
$x^2-x-2=x^2+2x-3$
$3x-1=0$

Since it is not a polynomial of degree 2, so it is not a  quadratic polynomial.
   

(iv)  ${(x – 3)(2x +1) = x(x +5)}$ 
 
 Solution  
$2x^2+x-6x-3=x^2+5x$   
$x^2-10x-3=0$
Since it is a polynomial of degree 2, so it is a quadratic polynomial.


(v) ${(2x – 1)(x – 3) = (x + 5)(x – 1)}$  
  Solution  
$2x^2-6x-x+3=x^2-x+5x-5$ 
$2x^2-7x+3 =x^2+4x-5 $
$x^2-11x+8=0$
Since it is a polynomial of degree 2, so it is a quadratic polynomial.


(vi) ${x^2 + 3x + 1 = (x -2)^2}$ 
 Solution  
$x^2+3x+1=x^2-4x+4$
$7x+3=0$
Since it is not a polynomial of degree 2, so it is not a  quadratic polynomial.

(vii) ${(x + 2)^3 = 2x (x^2 – 1)}$ 
  Solution 
$x^3+6x^2+12x+8=2x^3-2x$ 
$x^3-6x^2-14x-8=0$
Since the highest degree  of the ploynomial , is not a polynomial of degree 2,
 so it is not a  quadratic polynomial.

(viii) ${x^3 – 4x^2  – x + 1 = (x – 2)^3}$ 
  Solution 
  $x^3 – 4x^2  – x + 1= x^3-6x^2+12x-8$
  $2x^2-13x+7=0$
Since it is a polynomial of degree 2, so it is a quadratic polynomial.


2.   Represent the following situations in the form of quadratic equations :
 (i)   The area of a rectangular plot is 528 $m^2$. The length of the plot (in metres) is one 
 more than twice its breadth. We need to find the length and breadth of the plot.
 
 Solution
 Let the breadth of the rectangle be= $x$ $m$
 So, it lenght will be = $2x+1$ $m$
 Area of rectangle= length × breadth= $x$($2x+1$) =528
   $2x^2+x-528=0$
   $2x^2-32x+33x-528=0$
   $2x(x-16)+33(x-16)$
   $(x-16)(2x+33)$
   
   Therefore $x=16$.
Ans:   Therefore the length is $2x+1$= 33 $m$
   and breadth of rectangle = 16 $m$ 
   
(ii) The product of two consecutive positive integers is 306. We need to find the integers.
    
	Solution
	Let the 2 consectuive interger be $x$ and $x+1$
	Therefore $x(x+1)=306$
	i.e $x^2+x-306=0$
	$x^2+18x-17x-306=0$
	$x(x+18)-17(x+18)$
	$(x+18)(x-17)$
	
	i.e $x=17$
	
	Ans: Therefore the two consecutive number are 17 & 18.
	
(iii) Rohan’s mother is 26 years older than him. The product of their 
    ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.

	Solution 
	Let Rohan's age be $x$
	Hence, Rohan's mother age=$x+26$
	Given: Product of their ages (in years) 3 years from now will be 360.
	i.e (x+3)(x+29)=360
	     $x^2+29x+3x+87$=$360$
		 $x^2+32x-273=0$
		 
		 $x^2+39x-7x-273=0$
		 $x(x+39)-7(x+39)=0$
		 $(x+39)(x-7)=0$
		 $x=7$
	
	Ans:So, the Rohan's age is 7 year and his mother's age is 33 years.
  
  (iv)   A train travels a distance of 480 km at a uniform speed. If the speed had been
		8 km/h less, then it would have taken 3 hours more to cover the same distance.
		We need to find the speed of the train.
		
		Solution 
		Speed is given by the formula, $S=D/T$
		Let the speed of the train be $x$ km/h
		
		Now time taken in 1st case: $t_1=480/x$
		
		Now if the speed has been 8 km/h less, then new speed of the train is $x-8$ km/h
        Now time taken will be $t_1 +3$ .
		
		therefore $(t_1 +3)= 480/{x-8}$
		          $(480/x+3)= 480/{x-8}$
			i.e   $480(1/{x-8} - 1/x) = 3$
			      $160 ({x-x+8}/{x(x-8)})=1$
				  $1280=(x^2-8x)$
				  $x^2-8x - 1280 = 0$
				  $x^2-40x+32x-1280=0$
				  $x(x-40)+32(x-40) =0$
				  $(x-40)(x+32)=0$
				  
		i.e $x=40$ or $-32$
		Since the speed cannot be negative, so the speed of train is $40$ km/h