1.   Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:

(i)   ${2x^2 – 3x + 5 = 0 }$  
Solution
Recall from the chapter that , real roots exist if the discriminant of the quadratic equation,
i.e $b^2-4ac >= 0$
Therefore $b^2-4ac$ =$(-3)^2 -4(2)(5)$ =$9-40$ =$-31$
Since  $b^2-4ac <  0 $, so real roots does not exist.

(ii)  ${3x^2 –   4√3 x + 4 = 0}$ 
Solution
     $b^2-4ac$ = $(-4√3)^2-4(3)(4)$
	           =$48-48$
			   =0
	Since $b^2-4ac$ = 0, therefore real roots exist and are equal.
	and the roots are $-b/{2a}$ i.e $-(-4√3)/{2(3)}$ 
	= ${2√3}/3$

(iii)   ${2x^2 – 6x + 3 = 0}$

	Solution
	$b^2-4ac$ = $(-6)^2-4(2)(3)$
	          = $36-24$
		=$12$
	Since $b^2-4ac$ > 0, real roots exist. and the roots are,
	
	$x={-b±√{b^2-4ac}}/{2a}$
	  =${-(-6)± √12}/{2(2)}$
	  = ${+6 ±2√3}/4$
	$x$  =${3 ±√3}/2$
	

2.   Find the values of $k$ for each of the following quadratic equations, so that they have two equal roots.
(i)   ${2x^2 + kx + 3 = 0}$ 
   Solution
	Since they have equal roots, therefore their discriminant is zero.
	i.e $b^2-4ac = 0$
     $k^2-4(3)(2)=0$
     $k^2-24=0$
     $k^2=24$
	 $k=√24 $
	 $k=±2√6 $
	 
	 $k=±2√6 $ 
   

(ii)  ${kx (x – 2) + 6 = 0}$
	Solution
	${kx (x – 2) + 6 = 0}$ => $kx^2-2kx+6=0$
	Since they have equal roots, therefore their discriminant is zero.
	i.e $b^2-4ac = 0$
	$(-2k)^2-4(k)(6) = 0$
	$  4k^2-24k = 0$
	$  4k(k-6) = 0$$k=6$ 
	
Ans:	$k=6$ 
	
3.   Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 $m^2$? 
     If so, find its length and breadth.
	 Solution
	 Let the breadth = $x$ m
	 So, the length= $2x$ m
	 
	 Since the area= 800 $m^2$
               $x(2x)=800$	 
               $2x^2-800=0$
			   
    Since, for designing a rectangular mango groove, the roots of the eqn (1) shall be real.
	Hence $b^2-4ac >=0$
	      =$0^2-4(2)(-800)$
	     = $6400$ which is >0, hence the roots are real.
		
	$x={-b±√{b^2-4ac}}/{2a}$
	
	$x={0±√{6400}}/{2× 2}$	
	$x=80/4=20$
	
	Hence breadth of the rectangular mango groove = $20$ $m$
	and length of the rectangular mango groove = $40$ $m$
	
		 
4.   Is the following situation possible? If so, determine their present ages.
     The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.
	
    Solution
	Let the age of the friend be $x$ and $(20-x)$
     Age of the friends 4 years age was $x-4$ and $20-x-4$
     
     Product of their ages 4 years ago=48
                      $(x-4)(16-x) = 48 $                     
                      $16x-x^2 -64 +4x = 48 $					  
                      $-x^2 +20x -64 = 48 $					  
                      $ x^2 -20x +112 = 0 $	
    The roots of the quadratic equation will be real if $b^2-4ac >=0$	
	 i.e $b^2-4ac$ =$(-20)^2-4×1×112$
	               =$400-448$ 
				   =$-48$ which is < 0;
	
Ans:	Hence the situation is not possible. 
	 
5.   Is it possible to design a rectangular park of perimeter 80$m$ and area 400 $m^2$? If  so, find 
     its length and breadth.
	 
	 Solution
	 Let the length and breadth be $x$ and $y$.
	 Perimeter= $2(x+y)=80$ ...........(1)
	 Area     = $xy=400$ ...............(2)
	 
	 Substuting $y$ of eqn (2) in eqn (1)
	           => $2(x+400/x) = 80$
			   => $2x^2+800=80x$
			   => $2x^2 -80x+800=0$
			   
	Hence, $b^2-4ac$ = $(80)^2-4(2)(800)$
	                  =$6400- 6400$
					  =0
	Since the discriminant of the equation is zero, their roots are equal .
	and their roots are $-b/2a$ = $80/(2×2)$ = $20$.
	
	Ans:So, the rectangular park will be of equal sides of $20$ m each.