1) For what value of $k$, the roots of the equation $x^2+4x+k=0$ are real?.  

Solution
Since the roots are real, so $b^2   –  4ac  ≥  0$.
i.e $16-4k≥0 $
i.e $k≤4$
Ans: $k$ has all the real value ≤4 

2) Find the value of $k$ for which the roots of the equations $3x^2-10x+k=0$
are reciprocal of each other.Also find the roots.

Solution
Let the roots of the quadratic equation be  $α$ and $1/α$ 
Product of roots = $c/a$
Therefore $α × 1/α=k/3$
$k=3$

Hence, the quadratic equation will be $3x^2-10x+3=0$
Hence $3x^2-9x-x+3=0$
$3x(x-3)-1(x-3)=0$
$(x-3)(3x-1)=0$
Hence the roots are $3$ and $1/3$


Ans: k = 3 and roots are $3$ and $1/3$  

3)Solve the quadratic equation $6x^2-13x+6$ by method of completing the square.

Solution 
Since the coeficient of $x^2$ is not a perfect square, divide the equation 
throughout by 6. Therefore
$6x^2-13x+6$ = $x^2-13/6x+1$
$x^2-13/6x$ = $x^2-1/2(13/6)x - 1/2(13/6)x$
            = $x(x-13/12x) -13/12x$
Add and subract $(13/12)^2$, so the eqn become,
            = $x(x-13/12) -13/12x$ +$(13/12)^2$ - $(13/12)^2$ 
            = $x(x-13/12) -13/12(x-13/12)$ - $(13/12)^2$ 
			=$(x-13/12)^2-169/144$

$x^2-13/6x+1$ = $(x-13/12)^2-169/144 + 1$ 
              = $(x-13/12)^2-({169-144}/144)$ 
              = $(x-13/12)^2-({25}/144)$ 
              = $(x-13/12)^2-(5/12)^2$ 
			  =$(x-13/12+5/12)(x-13/12-5/12)$
			  =$(x-8/12)(x-18/12)$

Therefore the roots of the quadratic equation are:
$(x-8/12)=0$ => $x=3/2$
and $(x-18/12)=0$=> $x=2/3$


4)A water tank take 10 minute to fill completely. But a hole was observed at the bottom of the tank,
which can empty the tank in 25 minutes completely.How much time it will take to fill the tank now.
			  
Solution
Time taken to fill the tank now = $1/t=1/10-1/25$
                         = $(25-10)/{250}$
						 =${15}/250$
						 =$3/50$
						
Therefore time taken to fill the tank= $50/3$ = 16$2/3$ minutes.


5) Dhruv and Daniel has 100 marbles. Dhruv gave Daniel 20 marbles. Product of marbles
now they have is 2475. Find the number of marble each of them have originally. 

Solution
Let Dhruv has $x$ number of marble.
and Daniel has $(100-x)$ numbers of marble.

Dhruv gave Daniel 20 marbles.
Hence Dhruv has $(x-20)$ marbles,
and Daniel has $(120-x)$ marbles.

Given, Product of marbles now they have = 2475.
ie. $(x-20)(120-x)=2475$
    $120x-x^2-2400+20x=2475$
	$x^2-140x+4875=0$
	
	$x={-b±√{b^2-4ac}}/{2a}$
	$x={140±√{(-140)^2-4×4875}}/{2}$
	$x={140±√{19600-19500}}/{2}$
	$x={140±√{100}}/{2}$
	$x={140±{10}}/{2}$
	
Therefore $x={140+10}/2$ or $x={140-10}/2$
$x=75$ or $65$
Hence Dhruv has 75 or 65 marbles.
and Daniel has 25 or 35 marbles.

6) Find the discriminant of the quadratic equation $(x+5)^2=2(5x-3)$

Solution:
$x^2+10x+25=10x-6$
$x^2+31=0$
Discriminant = $b^2-4ac$
             = $0^2-4×1×31$
			 = -124

7) Find the value of $k$ for which the discriminant is 0 for the quadratic equation
$(x+2)^2=2(2x+k)$

Solution
$(x+2)^2=2(2x+k)$ 
$x^2+4x+4=4x+2k$
$x^2+4-2k=0$
Discriminant = $b^2-4ac$
0=$0^2-4×1×(4-2k)$	
$-16+8k=0$
$8k=16$
$k=16/8$
$k=2$


8) Solve the quadratic equation for $x$. 
		$x^2+({a}/{a+b}+{a+b}/{a})x + 1 = 0$

Solution 
		$x^2+({a}/{a+b}+{a+b}/{a})x + 1 = 0$
		$x^2+({a}/{a+b})x+({a+b}/{a})x + 1 = 0$
	Let 1 can be written as  $(a/{a+b})({a+b}/a)$
	
	Therefore 
	    $x^2+({a}/{a+b})x+({a+b}/{a})x + (a/{a+b})({a+b}/a) = 0$
		$x({x+a/{a+b}}) + ({a+b}/{a})(x+{a}/{a+b}) = 0$
		$({x+a/{a+b}})(x+{a+b}/{a}) = 0$
		
		Therefore $x= -a/{a+b}$ or $-(a+b)/a$
		

9) Find the value of $k$ for which the quadratic equations have two equal real roots.
$(x+2)^2=2(2x+k)$

Solution
Quadratic equation have two equal and real roots if  $b^2–4ac=0$
$(x+2)^2=2(2x+k)$ 
$x^2+4x+4=4x+2k$
$x^2+4-2k=0$
Discriminant = $b^2-4ac$
0=$0^2-4×1×(4-2k)$	
$-16+8k=0$
$8k=16$
$k=16/8$
$k=2$


10) If $√{(x^2+y^2)(a^2+b^2)}=(ax+by)$, prove that $x/a=y/b$
		
Solution 

Given $√{(x^2+y^2)(a^2+b^2)}=(ax+by)$
Squaring both the sides we get,
        ${(x^2+y^2)(a^2+b^2)}=(ax+by)^2$
		$x^2a^2+x^2b^2+y^2a^2+y^2b^2=x^2a^2+y^2b^2+2abxy$
		$x^2b^2+y^2a^2=2abxy$
		$(bx)^2-2(bx)(ay)+(ay)^2=0$
		$(bx-ay)^2 = 0$
		$bx=ay$
		$x/a=y/b$
		Hence Proved.
		

11) A shopkeeper sells a toy for Rupee 75 and gains as much precent as the cost price of the toy.
Find the cost price of the toy.

Solution 
Let the cost price of the toy be $x$ .
So gain percentage = $x%$

Gain amount = gain%× CP
            = $x/100 × x$

Since SP =CP + Gain Amount 
     $ 75  = x + x^2/100$
	 $7500 = 100x + x^2$
	 $x^2 +100x - 7500 = 0$
	 $x^2 +150x -50x - 7500 = 0$
	 $x(x+150) - 50(x+150)=0$
	 $(x+150)(x-50) = 0$
	 $x= -150$ or $50$
Since the Cost price cannot be negative, 
Therefore cost price =  Rs 50