1.   Find the roots of the following quadratic equations by factorisation:
(i)   ${x^2 – 3x – 10 = 0}$ 

Solution:    
		${x^2 – 3x – 10 = 0}$ => ${x^2 – 5x+ 2x  – 10 = 0}$ 
		$x(x-5)+2(x-5)$
		$(x-5)(x+2)$
		
		i.e $x=5$ and $-2$
                
 (ii)  ${2x^2 + x – 6 = 0}$
 
 Solution: 
		${2x^2 + x – 6 = 0}$ =>${2x^2 + 4x -3x – 6 = 0}$
		${2x(x+2)-3(x+2)}$
		$(x+2)(2x-3)$
		
		i.e $x=-2$ and $x=3/2$
 
 
(iii) ${ √2x^2  + 7 x + 5 √2 = 0}$ 
	Solution:
	${ √2x^2  + 7 x + 5 √2 = 0}$ => ${ √2x^2  + 2 x+ 5x + 5 √2 = 0}$

		${√2x(x+√2)+5(x+√2)   }$
		$(x+√2)(√2x+5)$
		
		i.e $x= - √2 $ or $x= -{5√2}/2$
              
(iv) ${ 2x^2 – x + 1/8= 0}$
     Solution:
(Remember: IN solving fraction type of problem, use multipication method and make it a whole number)
Multiply the equation by 8, therfore 
	 ${ 2x^2 – x + 1/8= 0}$ => ${ 16x^2-8x+1}$
	 $16x^2-4x-4x+1$
	 
	 $4x(4x-1)-1(4x-1)$
	 
	 $(4x-1)(4x-1)$
	 
	 i.e $x=1/4$ or $x=1/4$
	 
(v)   ${100 x^2 – 20x + 1 = 0}$ 
     Solution:
	 
	 ${100 x^2 – 20x + 1 = 0}$
	 $100 x^2 – 10x-10x + 1 = 0$
	 ${10(10x-1)-1(10x-1)}$
	 $(10x-1)(10x-1)$
	 
	 i.e $x=1/10$ and $x=1/10$
	 
3.   Find two numbers whose sum is 27 and product is 182.
    Solution:
	
	Let the number be $x$ and $(27-x)$
	Given  $x(27-x)=182$
	       $27x-x^2=182$
		   $x^2-27x+182$                  (	(182= 2× 13 ×7 ))

	       $x^2-14x -13x+182$
		   $x(x-14)-13(x-14)$
		   $(x-14)(x-13)
		   
	Ans:The two number are 13 and 14 
	
4.   Find two consecutive positive integers, sum of whose squares is 365.

 Solution:
     Let the two consecutive positive integer be $x$ and $x+1$
    Given: $x^2+(x+1)^2 = 365 $
	      i.e $x^2+x^2+2x+1 = 365$
		  $2x^2+2x-364=0$
		  $2x^2+27x-26x-364=0$
		  $x(2x+27)-13(2x+27)$
		  $(2x+27)(x-13)$
		  
	Since the numbe ris a postive interger, therfore x=13.
	
Ans:	SO, two positive integer are 13 & 14. 
		  

5.   The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm,
     find the other two sides.
	 Solution:
	 Let the base of the right triangle be $x$ $cm$.
	 Therefore its altitute = $(x-7)$ $cm$
	 
	 Therefore , for right triangle,
	  $H^2=B^2+L^2$
	  $(13)^2=x^2+(x-7)^2$
	  
	  $169=x^2 + x^2 -14x+49$
	  
	  $2x^2-14x-120=0$
	  
	  $2x^2-24x+10x-120=0$
	  
	  $2x(x-12)+10(x-12)=0$
	  
	  $(x-12)(2x+10)$
	  
	  Therefore base of right triangle $x=12$ $cm$ and its altitute is $5$ $cm$
	 

6.   A cottage industry produces a certain number of pottery articles in a day. It was observed 
     on a particular day that the cost of production of each article (in rupees) was 3 more than
	 twice the number of articles produced on that day. If the total cost of production on that 
	 day was ₹ 90, find the number of articles produced and the cost of each article.

Solution	 
    Let the number of pottery article produced in a day be $x$
	Given: Cost of production of each pottery article in a day = 3+2x
	Total cost of production in that particular day =  ₹ 90
	
	Therefore $x(3+2x)=90$
	i.e$3x+2x^2=90$
	$2x^2+3x-90=0$
	$2x^2+15x-12x-90=0$
	$x(2x+15)-6(2x+15) = 0$
	$(2x+15)(x-6)=0$