2.19 Quadratic Equations:
Do you find it
interesting to solve the following problem taken from Lilavati
(Shloka 71)?
In the epic
battle of Mahabharata, Arjuna takes out certain number of arrows.
He uses half of the arrows taken out to cut arrows of Karna, uses four
times the square root of number of arrows to target horses of Karna.
Uses 6 arrows to target Shalya, uses one each to target Chatra(Umbrella), flag and bow of Karna. Uses
the remaining one to target Karna.Then tell me the number of arrows taken
out by Arjuna.
Do you find it
interesting to solve the following real life problem?
Problem : Suppose you along with your friends had planned a picnic. You had
budgeted Rs.480 for food. But at the last moment 8 of your friends did not go
for the picnic. Because of their absence other members paid Rs.10 extra for
food. Find out how much each one paid finally?
We have learnt to solve problems like:
1. Find the side of a square if its perimeter is 60Meters.
Method: Let ’x’ be the side
of a square, then perimeter = 4x
Thus 4x =60
_{} x =15 Meters
A linear equation has only one
solution. The solution is called the root of the equation
2. If the area of a square is 25Sq.meters than what
is its side?
Method: Let ‘x’ be the side of a square. Then area
of the square = x^{2}
Thus x^{2} = 25 =5*5
_{} x=5 Meters
Since 25 = 5*5, x= 5 also satisfies the condition x^{2}
= 25. We can say that x = _{}5
are roots of the equation x^{2} = 25.
Because the side of a square can not be negative we
do not consider x = 5 Meters as a solution to the problem.
Definition:
An equation involving a variable of degree 2 is called a ‘quadratic
equation’
Note that x^{2} = 25 can also be expressed as x^{2}
 25 =0.
Note that the above equation has a variable only in
second degree and does not have variable in first degree (does not have terms
like bx)
Definition:
1.An equation of the type
ax^{2} +c = 0 where a and c are real numbers and a _{}0, is called a ‘pure quadratic equation’ One example is 3x^{2} 16=0
2. An equation of the type ax^{2} +bx+ c = 0
where a, b and c are real numbers and a
_{}0 and b _{}0, is called a ‘Adfected quadratic equation’ If b=0 then this equation becomes a pure
quadratic equation.
One example of an Adfected quadratic equation is 3x^{2}
5x16=0
Note : If ab =0 then either a=0 or b=0 or both a =0 and b=0
Example : Let us solve 3x^{2}
16=0
_{} 3x^{2} =16 (By
transposition)
x^{2} =16/3
_{}x = _{}= _{} _{}/_{}= _{} (4/_{})
2.19 Problem 1 : Solve x^{2}/2 –
3/4 = 29/4
Solution:
On transposition we
have
x^{2}/2 = 29/4+3/4 =
(29+3)/4 = 32/4 =8
_{} x^{2}
=16
_{} x = _{}4
2.19 Problem 2 : Solve (2m5)^{2}=
81
Solution:
(2m5)^{2}= 9^{2}
_{} 2m5 = _{}9
_{} 2m = _{}9 +5 ( On transposition)
_{} 2m = +9+5 =14 or 2m =
9+5 = 4
_{} m= 7 CxÀªÁ m= 2
Verification:
When m = 7 : LHS=(2m5)^{2}=(9)^{2}=81= RHS
When m =  2:
LHS=(2m5)^{2}=(45)^{2}=(45)^{2}=(9)^{2}=81=RHS
2.19 Problem 3 : If c^{2}= a^{2}+b^{2}
Solve for b. If a=8 and c=17 find the value of b
Solution:
Given c^{2}= a^{2}+b^{2}
_{} b^{2}= c^{2}a^{2}
_{} b = _{}(c^{2}a^{2})
Substituting
given values for a and c in the above equation we get
b = _{}(c^{2}a^{2})
= _{}(17^{2}8^{2})
= _{}(28964)
= _{}(225)
= _{}15
Verification:
When a=8 and b=15 we
have RHS= a^{2}+b^{2}=64+225 =289 = 17^{2}= c^{2}=LHS
2.19 Problem 4 : The volume of a
cylinder of radius ‘r’ and height ’h’ is
given by the formula Volume V = _{}r^{2}h
1. Solve for r.
2. Find the radius of the cylinder if Volume =176
and height =14
Solution:
Assume _{} = 22/7
Since V =_{}r^{2}h
r^{2}= V/_{}h
_{} r = _{}(V/_{}h)
It is given that V=176
and h = 14
V/_{}h = 176*7/(22*14)= 4
_{} r = _{}2
Since the radius can not be a negative number we
conclude that r=2 units
Verification:
Given_{} = 22/7, h =14 and arrived value for r=2: RHS= _{}r^{2}h= 22*4*14/7 = 22*4*2=176=V= LHS
2.19.1
Solving Adfected Quadratic equations by Factorisation method
In This method we first express the quadratic
equation as a product of 2 monomials and equate each of them to zero, and then
find values of the unknowns. This method requires a lot of practice and can be
mastered only over a period of time
2.19.1 Problem 1: Solve 6p^{2}=p
Solution:
This is equivalent to solving p^{2}+p6 = 0
( By transposition)
We need to express LHS in the form (x+a)(x+b) such that a+b =1 and ab = 6.
The factors of  6 are (1, 6), (1,6), (2,3), (2,3), (3,2),
(3,2)
We note that only a = 2 and b= 3 satisfy the conditions a+b=1 and ab =
6
_{}p^{2}+p6
= p^{2}+3p2p 6
= p(p+3) 2(p+3)  take out the common factor (p+3)
= (p+3)(p2)
Since p^{2}+p6 = 0
(p+3)(p2) = 0 (if product
of two terms is zero then one of the term has to be
zero)
This is possible if p+3 = 0 or p2 = 0
_{} p= 3 or p =2 are roots
of the given equation
Verification:
Let us put p=2 in the
given equation
LHS = 2^{2}+26
=4+26 = 0 =RHS. Similarly verify for p = 3
2.19.1 Problem 2: Solve 6 y^{2}+y 15 = 0.
Solution:
We need to express LHS in the form(ax+b)(cx+d)={
acx^{2} + x(ad+bc)+bd}such that ac=6, bd= 15 and ad+bc =1
By inspection it can be seen that a=3, c=2,b=5,d= 3 satisfy the given conditions
_{}6 y^{2}+y 15
= 6 y^{2}+10y 9y 15
= 2y(3y+5)3(3y+5) – take
out the common factor 3y+5
= (3y+5)(2y3)
Since 6 y^{2}+y 15 =0
(3y+5)(2y3) =0
This is possible if 3y+5 = 0 or 2y3 =0
_{} y = 5/3 or y =3/2 are roots of the given equation
Verification:
Let us put y=3/2 in the
given equation
LHS = 6*9/4 +3/2 15 =27/2+3/2 15 = (27+3)/2 – 15
= 0 = RHS
Similarly verify for y= 5/3
2.19.1 Problem 3: Solve 13m = 6(m^{2}+1)
Solution:
This is equivalent to
6m^{2}13m+6 =0
We need to express LHS in the form(ax+b)(cx+d)={
acx^{2} + x(ad+bc)+bd}such that ac=6, bd= 6 and ad+bc = 13
By inspection it can be seen that that a=3, c=2,b=5,d= 3 satisfy the given conditions
6m^{2}13m+6=0
= 6m^{2}9m 4m+6
= 3m(2m 3) 2(2m3) à take out the common
factor 2m3
= (2m3)(3m2)
Since 6m^{2}13m+6 =0
(2m3)(3m2)=0
This is possible if 2m3 = 0 or 3m2 =0
_{} m = 3/2 or m =2/3 are
roots of the given equation
Verification:
Let us put m=2/3 in the
given equation
LHS= 6*4/9 13*2/3 +6 = 8/3 26/3+6 =(826)/3 +6 = 0 =RHS. Similarly try for m= 3/2
2.19.1 Problem 4: Solve y^{2}2y+2 =0
Solution:
We need to express this equation in the
form(ax+b)(cx+d)={ acx^{2} + x(ad+bc)+bd}such that ac=1, bd= 2 and ad+bc = 2
Since ac=1 and bd=2, the possible values of a and c are: a=_{}1or c=_{}1 and (b=_{}2,d=_{}1) or (b=_{}1, d=_{}2).
We also notice that with whatever combination of a,
b, c, d, the condition ad+bc = 2 is not satisfied.
How do we
solve such equations?
While solving problems 2.19.1.1, 2.19.1.2, 2.19.1.3 we noticed that it is not
always easy to determine the factors.
Is it not logical to have a formula for finding
roots of such equations?
We shall explain the method by an example
Example : Let us solve 2x^{2}+3x+1
=0
No 
Step 
Explanation 
1 
x^{2}
+(3/2)x+ (1/2) =0 
Divide
both sides of the given equation by 2 
2 
x^{2}+(3/2)x=
(1/2) 
By
transposing (1/2) to RHS 
If
we can use the identity (x+a)^{2} = x^{2}+2ax+
a^{2} we could find a solution to the given equation. If the equation
in step 2 is compared with the above
identity, we can say 2ax
= (3/2)x and hence a =3/4 

3 
x^{2}+(3/2)x+ (3/4)^{2} = (1/2)+ (3/4)^{2}^{ } 
By
adding (3/4)^{2} to both sides of equation in step 2 
4 
LHS
of step 3= x^{2}
+2(3/4)x + (3/4)^{2}= [x+(3/4)]^{2} 
p^{2}+2pq+q^{2}
= (p+q)^{2} with p=x and q= 3/4 
5 
RHS
of step 3= (1/2)+ (3/4)^{2}^{
}=(1/2)+ (9/16)= (1/16) 

6 
[x+(3/4)]^{2}=(1/16) 
Step
4 and 5 
7 
(x+(3/4)) = _{}(1/4) 
Square
root of step 6 
8 
x
= (3/4) _{}(1/4) = (1/2) or 1 
Simplification 
We shall use the above method for solving generic
equation ax^{2} +bx+ c =0
Formula for finding
roots of the quadratic equation
Let us find the roots
of the Quadratic equation whose general form is ax^{2} +bx+ c =0, where
a, b and c are real numbers and a _{}0 and b _{}0.
No 
Step 
Explanation 
1 
x^{2}
+(bx/a)+ (c/a) =0 
Divide
both sides by ‘a’ 
2 
x^{2}
+(bx/a) = ( c/a) 
By
transposing c/a to RHS 
3 
x^{2} +(bx/a) + (b/2a)^{2}^{ }= ( c/a) + (b/2a)^{2}^{ } 
By
adding (b/2a)^{2} to both sides 
4 
LHS=
x^{2} +(bx/a) +
(b/2a)^{2}= [x+(b/2a)]^{2} 
p^{2}+2pq+q^{2}
= (p+q)^{2} with p=x and q= b/2a 
5 
RHS
= b^{2}/4a^{2} c/a= (b^{2}4ac)/ 4a^{2} 
By
having common denominator as 4a^{2} 
6 
[x+(b/2a)]^{2
}=(b^{2}4ac)/ 4a^{2} 
By
Step 4 and 5 as LHS=RHS 
7 
x+(b/2a) = _{} ((b^{2}4ac)/ 4a^{2})^{} =
_{} ((b^{2}4ac))/ 2a 
Take
square root of the last step 
8 
x = [b _{}_{} (b^{2}4ac)]/2a 
Transpose b/2a to RHS 
Therefore roots of the equation ax^{2} +bx+ c =0 are:
x = [b +_{}(b^{2}4ac)]/2a AND x = [b _{}(b^{2}4ac)]/2a
Note : This formula called quadratic
formula was first given by the Indian mathematician
Sridharacharya
(1025AD) The formula is given in Lilavati
also.(Shloka 67)
2.19.1 Problem 5: Solve 4x^{2}+8x+4 = 0
Solution:
Here we have a =4, b=8,
c =4
_{} b^{2}4ac = 64 – 4*4*4 = 0
_{} _{}(b^{2}4ac)
= _{}(0)
= 0
There fore as per the
formula roots are
p = [b +_{}]/2a =(8+0)/8 =  1 or
p = [b _{}]/2a = (80)/8 =  1
Here the roots are same =  1
Alternatively, note the given equation is
equivalent to 4(x^{2}+2x+1) = 4(x+1)(x+1)
which again suggests that roots are 1.
2.19.1 Problem 6: Solve p^{2}+p6 = 0(Repetition of problem
2.19.1.1 solved earlier)
Solution:
This equation is of the
form ax^{2} +bx+ c =0
Here we have a =1, b=1, c =6
_{} b^{2}4ac = 1 – 4*1*(6) = 25
_{} _{} = _{}(25)
= _{}5
As per the formula,
roots are
p = [b +_{}]/2a =(1+5)/2 = 2 or
p = [b _{}]/2a = (15)/2 = 3
These are the roots we got earlier
2.19.1 Problem 7: Solve 6y^{2}+y
15 = 0(Repetition of problem 2.19.1.2 solved earlier) and then factorise.
Solution:
This equation is of the
form ax^{2} +bx+ c =0
Here we have a=6, b=1, c= 15
_{} b^{2}4ac = 1 – 4*6*(15) = 361
_{} _{}(b^{2}4ac)
= _{}(361)
= _{}19
As per the formula,
roots are
y = [b +_{}]/2a =(1+19)/12 = 18/12= 3/2 or
y = [b _{}]/2a = (119)/12 = 20/12 = 5/3
These are the roots we got earlier
Since 3/2 and 5/3 are roots of the given equation,
(y3/2)(y+5/3) are factors of the given equation
Note (y3/2)(y+5/3) =
(2y3)(3y+5)/6
_{}6y^{2}+y 15 = (2y3)(3y+5)
Exercise: Solve example 2.19.1.3
using the formula method
2.19.1 Problem 8: Solve y^{2}2y+2 =0(Repetition of problem 2.19.1.4
which was not solved earlier)
Solution:
This equation is of the
form ax^{2} +bx+ c =0
Here we have a=1, b=2, c=2
_{} b^{2}4ac = 4 – 4*1*2 = 4
_{} _{}(b^{2}4ac)
= _{}(4)
= 2_{}
As per the formula,
roots are
y = [b +_{}]/2a =(2+2_{})/2 = 1+_{} or
y = [b _{}]/2a = (22_{})/2 = 1_{}
Because the
root contained non real number we could not factorize in problem 2.19.1 Problem
4
Verification:
Let us put y= 1+_{} in the given equation
y^{2}2y+2 = (1+_{})^{2} 2(1+_{}) +2 (Use the formula (a+b)^{2} =a^{2}+b^{2}+2ab
to expand (1+_{})^{2})
= [1 +(1) +2_{} ] +[2 2_{}] +2
=
11 +2_{} 2 2_{}+2 = 0 = RHS.
Similarly you can verify for other root= 1_{}
2.19.1 Problem 9: Solve 2(3y1)/(4y3) =
5y/(y+2) 2
Solution:
RHS = [5y 2(y+2)]/(y+2) = (3y4)/(y+2)
_{} We need to solve
2(3y1)/(4y3) = (3y4)/(y+2)
On cross multiplication we get 2(3y1)*(y+2) =
(3y4)*(4y3)
i, e 2(3y^{2}+6y –y 2) =
12y^{2}9y 16y+12
_{} 6y^{2}+10y 4 =
12y^{2}25y +12(By simplifying after transposing all terms from LHS to
RHS we get:)
0 = 6y^{2}35y +16: 6y^{2}35y +16=0
This equation is of the form ax^{2} +bx+ c
=0
Here we have a=6, b=35, c= 16
_{} b^{2}4ac = 1225 – 4*6*16 = 1225384 = 841
_{} _{}(b^{2}4ac)
= _{}(841)
= 29
As per the formula,
roots are
y = [b +_{}]/2a =(35+29)/12 = 16/3 or \
y = [b _{}]/2a = (3529)/12 = 1/2
Verification:
Substituting these
values in the equation
it can be seen that
LHS=RHS
2.19.1 Problem 10: Solve (y1)(5y+6) /(y3)
= (y4)(5y+6)/(y2)
Solution:
On cross multiplication
in the equation we get
(y1)(5y+6)(y2) = (y4)(5y+6)(y3) on
expanding terms on both LHS and RHS we
get
LHS = (5 y^{2}+6y5y6)(y2)
= (5 y^{2}+y6)(y2)
= 5 y^{3}+ y^{2}6y 10 y^{2}2y+12
=5 y^{3} 9y^{2}8y+12
RHS
=
(5y^{2}+6y20y24)(y3)
= (5y^{2}14y 24)(y3)
= 5y^{3}14 y^{2}24y 15y^{2}+42y+72
= 5y^{3}29y^{2}+18y+72
Since it is given that LHS=RHS we have
5 y^{3} 9y^{2}8y+12= 5y^{3}29y^{2}+18y+72.
(On transposing all the terms from RHS to LHS we get:)
5 y^{3} 9y^{2}8y+12(5y^{3}29y^{2}+18y+72)
=0(On simplification we get)
20y^{2}26y60 = 0 ( By
taking out 2 as a common factor)
10y^{2}13y30 = 0
This equation is of the form ax^{2} +bx+ c
=0
Here we have a=10, b=13, c= 30
_{} b^{2}4ac = 169 – 4*10*(30) = 169+1200 = 1369
_{} _{}(b^{2}4ac)
= _{}(1369)
= 37
As per the formula,
roots are
y = [b +_{}]/2a =(13+37)/20 =
50/20 = 5/2 or
y = [b _{}]/2a = (1337)/20 = 24/20 = 6/5
Verification:
Substituting these
values in the equation
it can be seen that
LHS=RHS
Alternative method of solving this
problem:
Since (5y+6) is common factor for both sides in the
given equation, we have two alternatives:
(1). When 5y+6 = 0:
Then we have 5y= 6 I.e. y = 6/5
_{} y = 6/5 is a solution to the
given problem à(1)
(2) When 5y+6 _{} 0 we can divide both
sides of the given equation by 5y+6 then we get
[(y1)/(y3)] =[(y4)/(y2)]
: By cross multiplication we get
(y1)(y2) = (y4)(y3)
i,e y^{2}2yy+2 = y^{2}3y4y+12
i,e y^{2}3y+2 = y^{2}7y+12: (On
transposition we get)
i,e y^{2}3y+2( y^{2}7y+12)=0
i,e y^{2}3y+2y^{2}+7y12=0
i,e 4y10=0
i,e 4y=10 or y=10/4 =5/2 à(2)
From (1) and (2) we conclude that 5/2 and 6/5 are
roots of the given equation
2.19.1 Problem 11: Solve y/(y+1) + (y+1)/y = 25/12
Solution:
On simplifying LHS we
get
[y*y +(y+1)(y+1)]/[y(y+1)]
= (y^{2}+y^{2}+2y+1)/( y^{2}+y)
Since LHS = RHS we get
(y^{2}+y^{2}+2y+1)/( y^{2}+y) = 25/12 (On cross multiplication we get)
12(y^{2}+y^{2}+2y+1) = 25( y^{2}+y)
_{} 24y^{2}+24y+12
= 25y^{2}+25y. On transposing LHS to RHS we get
0 = y^{2}+y12
This equation is of the form ax^{2}
+bx+ c =0
Here we have a=1, b=1, c= 12
_{} b^{2}4ac = 1 – 4*1*(12) = 1+48 = 49
_{} _{} = _{}(49)
= _{}7As
per the formula, roots are
y = [b +_{})]/2a =(1+7)/2 = 3 or
y = [b _{})]/2a = (17)/2 =  4
Verification:
Substituting these
values in the equation it can be seen that LHS=RHS
2.19.1 Problem 12 : Solve (3x^{2}5x+2)
(3x^{2}5x2)=21
Solution:
1. Let 3x^{2}5x
= y then solve for y in (y+2) (y2) =21
2.
Substituting value for y in the equation 3x^{2}5x = y solve for
x.
Answer: x =  (5) _{} _{}(25+60)/2*3
= 5 _{} _{}(85)/6
2.19.1 Problem 13 (Problem given at the
start of this section): Suppose you along with your friends had planned a picnic.
You had budgeted Rs.480 for food. But at the last moment 8 of your friends did
not go for the picnic. Because of their absence other members paid Rs.10 extra
for food. Find out how much each one paid finally?
Solution:
Let ‘x’ be the number
of people who were supposed to go to picnic.
Therefore the food bill
per head = 480/x
Since 8 did not join
finally only (x8) people went for the picnic
_{} The revised food bill
per head = 480/(x8)
This is given to be Rs
10 more than what was planned earlier
_{} The new rate = old rate
+10
So we have 480/(x8) =
480/x + 10
After simplifying RHS
we get
480/(x8) =
(480+10x)/x. (On cross multiplication we get)
480x = (480+10x)(x8) (On expanding RHS we get )
RHS= 480x 480*8
+10x*x80x
= 480x  3840+ 10x^{2}80x = 10x^{2}+400x3840
_{}0 =10x^{2}+400x3840480x. (By transposing 480x to RHS)
I.e.
0 =10x^{2}80x3840. Dividing both the sides of this equation by 10 we
get
x^{2}8x384 =0
This equation is of the form ax^{2} +bx+ c
=0
Here we have a=1, b= 8, c= 384
_{} b^{2}4ac = 64 – 4*1*(384) = 64 +1536 =1600
_{} _{} = _{}(160000)
= 40
As per the formula, roots are
x = [_{}] =(8+40)/2 = 24
or
x = [b _{}]/2a = (840)/2 = 16
Since number of people can not be negative, the
correct solution is 24
Thus 24 friends
had planned to go out for a picnic
Therefore the revised food bill per head =(_{})=30 Rs
Verification:
Since 24 people had
planned to go out for a picnic. The cost of food per head which was planned,
was = 480/24 = Rs.20
Since 8 did not go,
only 16 went for picnic
Therefore the revised
cost of food is 480/16 = Rs 30 which is Rs 10 more than what was planned. This
result matches with what is given in the problem.
2.19.1 Problem 14: Hypotenuse of a right angled triangle is 20mts. If
the difference between lengths of other 2 sides is 4mts. Find
the length of the sides
Solution:
If x and y are the sides of a
Right angled triangle then by Pythagoras theorem we know that (Hypotenuse)^{2} = x^{2}+
y^{2} .It is given that hypotenuse =20_{}20^{2} = x^{2}+ y^{2} =======è (1) Since we are given that xy = 4:
We have x= 4+y. Substituting this value of x in equation (1) and then
expanding we have 400 = x^{2}+ y^{2} =(4+y)^{2}+
y^{2} = (16+8y+ y^{2})+ y^{2}=16+8y+ 2y^{2} .
On transposing terms from LHS to
RHS we have 0 = 2y^{2}+8y384.This equation is of the form ax^{2}
+bx+ c =0 Here
we have a=2, b= 8, c= 384 _{} b^{2}4ac
= 64 – 4*2*(384) = 64+3072 =3136 _{} _{} =_{}(3136)
= 56.As per the formula, roots are y = [b +_{})]/2a =(8+56)/4
= 12 or y
= [b _{})]/2a = (856)/4 = 16 Since
the side of a triangle can not be a negative number, the correct answer for
one side y =12mts and hence another side is16 mts(x=4+y) 

Verification:
(side)^{2}+
(side)^{2} = 12^{2}+ 16^{2} = 144+ 256 = 400 =20^{2}
.Therefore hypotenuse=20 which is as given in the problem
2.19.1 Problem 15: The distance between 2 cities is 1200km. A super
fast train runs between these 2 cities. When the speed is increased by 30km/hr
from its initial speed the journey time reduces by 2 hours. Find the initial
speed of the train.
Solution:
Let x be the
initial speed. Therefore time taken = 1200/x
If speed is increased
by 30 km/hr then the revised time taken is 1200/(x+30).
It is given that the
new time taken is 2 hours less than the original time
_{} 1200/x1200/(x+30) = 2
Exercise : Apply the formula to
get the correct answer x=120
Verification:
1200/120 – 1200/150 =
108 =2 which is as given in the problem.
2.19.1 Problem 16: A sailor operates a motor
boat between 2 ports which are 8 km apart. . He covers the journey (both ways)
between 2 ports in 1hour 40minutes.If the speed of stream is 2km per hour. Find
out the speed of boat in still water.
Note that, he has to
sail the boat along with the stream in one way (reduces the journey time).On
the return journey he has to sail the boat against the stream (The journey time
increases)
Solution:
Let x be the speed of
the boat.
We are given:
Total time taken to
cover up and down = 1hr 40mins = 100/60 hour = 5/3 hours
Distance between port =
8km
The speed of stream is
2km/hr
_{}Time taken to row down = 8/x+2 (Speed is the combined
speed of stream and boat)
_{}Time taken to row up = 8/x2(Speed reduces by the speed of
stream)
_{}Total time taken = 8/(x2) + 8/(x+2) which is given to be
5/3
Thus the equation to be
solved is 8/(x2) + 8/(x+2) = 5/3
Exercise: Apply the formula to get the
correct answer x =10
Verification:
Total time taken = 8/(102) + 8/(10+2) = 8/8 + 8/12 = 1+2/3 = 5/3 which is the time given in the problem
2.19.1 Problem 17: A plane left 30 minutes
later than the scheduled time. In order to reach the destination 1500 km away
it has to increase the speed by 250km/hr from its regular speed. Find the
regular speed and its normal journey time.
Solution:
Let x be the regular
speed of the plane
The distance to be covered is 1500km
_{} Normal journey time =
distance/speed =
1500/x
Since the plane started late by half an
hour, the speed was increased to cover
the distance so that still it reached on
time.
Thus the time available
for the plane to cover is = (1500/x) 1/2
During this time
it still flew 1500 km with the speed of
(x+250)
_{} distance
= reduced journey time*new speed
I.e. 1500 = {(1500/x)
1/2}*(x+250) = (30002x)*(x+250)/2x
I.e. 3000x = (3000x)(x+250) ( By cross multiplication)
I.e. 3000x = 3000x x^{2}+750000250x
I.e. x^{2}750000+250x
=0
Apply the formula to get_{} = 1750
_{} roots
are :
x = [b _{})]/2a = (250 _{}1750)/2
Which gives x = 750 or
x =1000
Since the plane can not
fly in a negative speed the solution has to be x = 750km/hr
_{} Normal journey time =
1500/750 = 2hr
Verification:
When the speed is
increased by 250km/hr the new speed becomes 1000km/hr
_{} The time taken to
cover 1500km = 1500/1000 = 1.5 Hours which is less than the normal flying hours
by half an hour.
Since the plane left half
an hour late, with the increased still it could reach on time. Hence our
solution is correct.
2.19.1 Problem 18: O Girl, out of group of
swans, 7/2 times the square root of the number are playing on the shore of a
tank, the remaining two are fighting among themselves in the water. Find the
total number of swans (Bhaskara 1114AD : ‘Leelavati :shloka’ 68)
Solution:
Let x be the total
number of swans
The number of swans
playing on the shore of tank = (7/2)_{}
The number of swans
fighting in water = 2
Thus we are required to
solve the equation
x= (7/2)_{}+2
On solving we find the
roots as 1/4 or 16
Since 1/4 is not
feasible, the number of swans has to be 16
Verification:
Note 16 = 14+2 = (7/2) _{}+2 which is as given in the problem and hence our
solution is correct.
2.19.1 Problem 19: In the epic battle of Mahabharata,
Arjuna takes out certain number of arrows. He uses half of the arrows
taken out to cut arrows of Karna, uses four times the square root of
number of arrows to target horses of Karna. Uses 6 arrows to target Shalya,
uses one each to target Chatra(Umbrella), flag
and bow of Karna. Uses the remaining one to target Karna.Then
tell me the number of arrows taken out by Arjuna.(Lilavati Shloka 71)
Solution:
Let x be the total
number of arrows.
No 
Target 
How many 
1 
Arrows of Karna 
(x/2) 
2 
Horses of Karna 
4 _{} 
3 
Shalya 
6 
4 
Chatra, flag and bow of Karna 
(1+1+1)
=3 
5 
Karn(Remaining) 
1 
_{}x = (x/2)+ 4 _{}+6+3+1
_{} x –(x/2)10 = 4 _{}
_{} (x/2)10
= 4 _{}
_{} (x20)
= 8 _{}
_{} x^{2}40x+400
= 64x à (a+b)^{2}Formula).
_{} x^{2}104x+400
=0
_{} (x100)*(x4)
=0
_{} x=100 Or
x=4
Number of arrows has to
be more than 6 as Arjuna uses 6 to target Shalya. Hence the
number of arrows used by Arjuna is 100
Verification:
100= 50+40+6+3+1
2.19.1 Problem 20: In a forest, square of
3 less than 1/5^{th} of the group of monkeys went inside a cave. If the remaining one went
up a tree, find the total number of monkeys in the group (Bhaskaracharya : Bijaganita)
Solution:
Let x be the total
number of monkeys
in the group.
No 
To where? 
How many 
1 
Cave 
{(x/5)3}^{2} 
2 
Remaining 
1 
_{}{(x/5)3}^{2}+1 =x
_{}(x^{2}/25) –(6x/5)+9+1=x
_{}(x^{2}/25) –(11x/5)+10=0
_{}x^{2}–55x+250=0
_{}(x50)*(x5) =0
_{} x=50 Or x=5. It can not be 5, as (x/5)3 can not be negative.
Verification:
50= (103)^{2}+1=
49+1,
2.19.1 Problem 21: Solve 12( x^{2}+ 1/ x^{2}) 56(x+1/x) = 89
Solution:
The given equation is
12( x^{2}+ 1/ x^{2}) 56(x+1/x) +89 =0
We know (x+1/x)^{2}^{ } = x^{2}+ 1/ x^{2}+2
_{} x^{2}+
1/ x^{2}=(x+1/x)^{2 }2
Substituting LHS value
of the above equation in the given equation we get
12{(x+1/x)^{2}^{ }2} 56(x+1/x) +89=0
I.e. 12(x+1/x)^{2}^{ }24 56(x+1/x) +89=0
I.e. 12(x+1/x)^{2}^{ } 56(x+1/x) +65=0
Let (x+1/x)= y
_{}We are required to solve 12y^{2}56y+65 =0
By applying the formula
we find that roots of this equation are y=5/2 or y=13/6
Case 1 :When y= 5/2: By substituting
value of y we get
_{}(x+1/x)= 5/2
I.e. (x^{2}+1)/x
= 5/2
I.e. 2(x^{2}+1)
= 5x
I.e. 2x^{2}5x+2
=0
By applying formula we
get the roots of this equation as 2, 1/2
Case 2 :When y= 13/6: By
substituting value of y we get
_{}(x+1/x)= 13/6
I.e. (x^{2}+1)/x
= 13/6
I.e. 6(x^{2}+1)
= 13x
I.e. 6x^{2}13x+6
=0
By applying formula we
get the roots of this equation as 2/3, 3/2
So the roots of the
given equation are {2, 1/2, 2/3, 3/2}
2.19.2
Nature of roots of a Quadratic equation
Observations : Have you observed the values of b^{2}4ac
in solving the problems?
In problem 2.19.1 .5,
we have seen that b^{2}4ac = 0 and roots are same
In problem 2.19.1 .8, we have seen that b^{2}4ac
<0 and roots are not real numbers
In all other examples
we have seen that b^{2}4ac > 0 and roots are real numbers
The expression b^{2}4ac
is called ‘discriminant’
and is denoted by _{}(called delta)
We conclude the
following:

Value
of Discriminant(b^{2}4ac) =_{} 
Nature
of roots=[b _{}_{}]/2a 
1 
_{} = 0 
Roots are real and equal 
2 
_{} >0 (Positive) 
Roots are real and distinct(not
equal) 
3 
_{} <0 (Negative) 
Roots are imaginary(not real)
and distinct 
2.19.2 Problem 1: For what positive
values of ‘m’, the roots of mk^{2}3k+1 =0 are equal, (real and
distinct) and (imaginary and distinct)
Solution:
Here we have a=m, b=
3, c= 1
_{} b^{2}4ac = 9 – 4m
1. Roots are equal when b^{2}4ac =0
(I.e. 94m =0, i.e. m = 9/4)
2. Roots are real and not equal when b^{2}4ac
>0
(I.e. 94m >0, i.e. 9 >4m, i.e. m < 9/4)
3. Roots are imaginary and not equal when b^{2}4ac
<0
(I.e. 94m <0, i.e. 9 <4m, i.e. m > 9/4)
2.19.2 Problem 2: For what values of ‘m’,
the roots of r^{2}(m+1)r +4 =0 are equal,
(real and distinct) and (imaginary and distinct)
Solution:
Here we have a=1, b= (m+1), c= 4
_{} b^{2}4ac = (m+1) ^{2}16
= [(m+1)+4]*[(m+1)4] ===> By factorization
= (m+5)(m3)
1. Roots are equal when b^{2}4ac =0
(i.e. (m+5)= 0 or (m3)=0
i.e. m=5 or m=3
2. Roots are real and not equal when b^{2}4ac
>0
(i.e. (m+5)(m3) >0) (Note
that when the product of 2 terms is +ve then both terms have to be +ve or both
terms have to be –ve)
This is possible in two cases
Case 1: both
m+5 > 0 and m3>0
I.e. m> 5 and m>3: This is possible only if
m>3
Case 2: both
m+5 < 0 and m3 <0
I.e. m< 5 and m<3: This is possible only if m
<5
3. Roots are imaginary and not equal when b^{2}4ac
<0
(i.e. (m+5)(m3) <0) (Note
that when the product of 2 terms is ve then one of the term is +ve and other
term is ve)
This is possible in two cases
Case 1: m+5 <0 and m3>0
I.e. m< 5 and m>3: This is not possible at all
Case 2: m+5 > 0 and m3 <0
I.e. m> 5 and m<3: That is when value of m is between 5 and 3
The above findings can be represented on number
line as follows.
2.19.2 Problem 3: Find value of ‘p’ for which (p+1) n^{2}+2(p+3)n +(p+8) =0 has equal roots
Solution:
This equation is of the
form ax^{2} +bx+ c =0
Here we have a=(p+1), b=
2p+6, c= p+8
_{} b^{2}4ac = (2p+6)^{2} – 4*(p+1)(p+8) = (4p^{2}+24p+36)
4(p^{2}+8p+p+8)= 4p^{2}+24p+36 4p^{2}36p32
=12p+4
If the roots are to be equal then b^{2}4ac
=0
I.e. 12p+4 = 0
I.e. p=1/3
As per the formula, roots for p=1/3
are
n = [b _{}]/2a =[2(p+3)_{}0)
]/2(p+1) =  (p+3)/(p+1)
=
 (10/3)/(4/3) = 5/2
Verification:
Substituting n
= 5/2 in the equation we get
(p+1) n^{2}+2(p+3)n +(p+8)
= 25(p+1)/4 5(p+3) +(p+8)
= 25(p+1)/4 4p 7 ( By having
4 as common denominator, we get)
= (25p+2516p28)/4
= (9p3)/4 (By
substituting p = 1/3 we get
=0/4 = 0 = RHS of the
given equation
2.19.2 Problem 4: Find value of ‘p’ for
which (3p+1) c^{2}+2(p+1)c +p =0 has equal roots
Solution:
This equation is of the
form ax^{2} +bx+ c =0
Here we have a=(3p+1), b=
2p+2, c= p
_{} b^{2}4ac = (2p+2)^{2} – 4*(3p+1)p = (4p^{2}+4+8p)
4(3p^{2}+p)= 4p^{2}+4+8p
12p^{2}4p = 8p^{2}+4p+4 = 
4(2p^{2}p1)
If the roots are to be equal then b^{2}4ac
=0
I.e. 2p^{2}p1 = 0
LHS = 2p^{2}2p+p1 = 2p(p1)+(p1)
= (p1)(2p+1)
Since 2p^{2}p1 = 0 we have
(p1)(2p+1) = 0
_{} p=1 or p= 1/2 are the answers
NOTE: To find roots of 2p^{2}p1 = 0 we
used factorisation method.
As per the formula, roots with p=1 is
c = [b +_{}]/2a =[2p2 _{}0)
]/2(3p+1) =  4/8 =
1/2
With p = 1/2 we get another value for c
NOTE: In the
above problem we could use the formula twice to work out the example.
Verification:
Substituting c = 1/2 in
the equation we get
(3p+1) c^{2}+2(p+1)c +p
= (3p+1)/4+2(p+1)(1/2) +p
=(3p+1)/4 –(p+1) +p
=(3p+1)/4 1 (By having 4
as common denominator we get)
= [(3p+1) 4]/4 (By
substituting p =1 we get
= 0/4 = 0=RHS of given
equation
Exercise : Verify that p =
1/2 gives equal roots for (3p+1) c^{2}+2(p+1)c
+p =0
2.19.2 Problem 5: Find value of ‘p’ for
which 2y^{2}py +1 =0 has equal roots
Solution:
This equation is of the
form ax^{2} +bx+ c =0
Here we have a=2, b= p, c= 1
_{} b^{2}4ac = p^{2} 8
If the roots are to be equal then b^{2}4ac
=0
I.e. p^{2} = 8 :
I.e. p = _{} 2_{}
Exercise: Verify that this value of p gives equal roots to the given
equation
2.19.3 Relationship
between roots and coefficients:
Let ‘m’ and ‘n’ be the roots of quadratic
equation of the form ax^{2} +bx+ c =0
_{} (xm)(xn) = 0
We also have seen that
the roots of this equation are or
x = [b +_{}]/2a or x = [b _{}]/2a
_{} m = [b +_{}]/2a
n = [b _{}]/2a
_{} m+n = [b +_{}]/2a + [b _{}]/2a
= 2b/2a = b/a
mn = [b +_{}]/2a * [b _{}]/2a (By applying formula
for (a+b)(ab) we get
= [ (b)^{2} {_{}}^{2}] /4a^{2}
= [b^{2} (b^{2}4ac) ] /4a^{2}
= 4ac/4a^{2}
= c/a
We conclude:
1) Sum of the roots of
a quadratic equation = b/a
2) The product of roots
of a quadratic equation = c/a
2.19.3 Problem 1: Find the sum and product of roots of x^{2} +(ab)x+ (a+b) =0
Solution:
This equation is of the
form ax^{2} +bx+ c =0
Here we have a=1, b= ab, c= (a+b)
_{}m+n = b/a = ab/1 = ab
mn =c/a =(a+b)/1 = (a+b)
2.19.3 Problem 2 Find the sum and product of roots of pr^{2}
= r5
Solution:
This is equivalent to
pr^{2} –r+5= 0
This equation is of the form a x^{2} +bx+ c
=0
Here we have a=p, b= 1, c= 5
_{}m+n = b/a = 1/p
mn =c/a = 5/p
2.19.4 Formation of
equation with given roots
If ‘m’ and ‘n’ be the roots of the quadratic
equation ax^{2} +bx+ c =0
Then we know (xm)(xn) = 0
But (xm)(xn)
=x(xn)m(xn)
= x^{2} –xn –mx +mn
= x^{2 }–x(n+m)
+mn
= x^{2 }–(n+m)x
+mn
The general format is
x^{2}^{ }–(sum of roots)x +(product of
roots) =0
2.19.4 Problem 1: If ‘p’ and ‘q’ are the roots of the equation 2a^{2}4a+1=0
find the value of (p+q)^{2}+4pq
and p^{3} +q^{3} and also form the equation whose roots are p^{3}
and q^{3}
Solution:
This equation is of the
form ax^{2}+bx+ c =0
Here we have a=2, b= 4, c= 1
_{}p+q = b/a = 4/2 =2
pq =c/a =1/2
_{} (p+q)^{2}+4pq=4+2 =6
We know the general
formula for a^{3}+b^{3}= (a+b) (a^{2}+b^{2}ab)
_{}p^{3} +q^{3}
= (p+q)( p^{2}+q^{2}pq)
= (p+q)[( p^{2}+q^{2}+2pq) 3pq)]
= (p+q)[( p+q)^{2}3pq]
=2*[43/2] =5 (By
substituting vales for (p+q) and pq )
We are also required to form an equation whose
roots are p^{3} and q^{3}
Sum of roots = p^{3} +q^{3} =5( we had just calculated above)
Product of roots = p^{3}*q^{3} = (pq)^{3} =(1/2)^{3}
=1/8
_{} The desired equation
is
x^{2}(sum of roots)x+ (product
of roots)= 0
I.e. x^{2}5x+ 1/8= 0 (by multiplying terms
by 8 we get)
I.e. 8x^{2}40x+1=0
2.19.4 Problem 2: Form a quadratic
equation whose roots are p/q and q/p
Solution:
We are given m =p/q,
n=q/p
_{}m+n = p/q+q/p = (p^{2}+q^{2})/pq:
mn = p/q*q/p =1
_{}The standard form is
x^{2 }–(n+m)x +mn= 0
I.e. x^{2 }–(p^{2}+q^{2})x/pq
+1 = 0
I.e. (pqx^{2 }–(p^{2}+q^{2})x
+pq)/pq =0( Have pq as common denominator)
I.e. pqx^{2 }–(p^{2}+q^{2})x +pq=0
2.19.4 Problem 3: If one root of the equation x^{2}+px+q=0 is 3 times the other prove that 3p^{2}=16q
Solution:
This equation is of the
form ax^{2}+bx+ c =0
Here we have a=1,b=p,c=q
Let m and n be the roots of the equation.
_{}m+n = b/a =  p and
mn=c/a = q
It is given that one of the root
is 3 times another. So let m =3n
_{} p =  (m+n) =(3n+n)= 4n and q =mn=3n*n = 3n^{2}
_{} 3p^{2}= 3(4n)^{2}= 48 n^{2}=16*3n^{2}^{ }= 16q(_{}3n^{2}=q)
2.19.4 Problem 4: Find the value of ‘p’ so that the equation 4x^{2}8px+9=0 has roots whose difference is 4
Solution:
This equation is of the
form ax^{2}+bx+ c =0
Here we have a=4,b=8p,c=9
Let m and n be the roots of the equation
_{}
1) m+n = b/a = 8p/4=2p
===è(1)
2) mn= c/a = 9/4
===è(2)
Since the difference between roots is 4 let n=m+4
Substituting this value in (1) we get
m+n = m+m+4 = 2p: i,e 2m=
2p4: i,e m=p2 à(3)
By substituting
n= m+4 in (2) we get
m(m+4) =9/4
I.e. m^{2}+4m  9/4 =0
I.e. (p2)^{2}+4(p2)  9/4 =0(_{}m=p2 as per (3))
I.e. p^{2}4p+4 +4(p2)  9/4 =0 (By
expanding (p2)^{2} using formula )
I.e. p^{2}4p+4 +4p8  9/4 =0
I.e. p^{2}4  9/4 =0
I.e. p^{2}25/4 =0
I.e. p^{2}= 25/4
p = _{} 5/2
Verification: Substitute value of p (5/2) in
the given equation we get
4x^{2}8px+9=0
i.e. 4x^{2}8*(5/2)x+9=0
i.e. 4x^{2}+20x+9=0 This is of the form
ax^{2}+bx+c=0 where a=4, b=20,
c=9
_{} b^{2}4ac = 400 – 4*4*9 = 400144 =256
_{} _{} = _{}(256)
= 16
As per the formula, roots are
x = [b +_{}]/2a =(20+16)/8 =
4/8
x = [b _{}]/2a = (2016)/8 = 36/8
Notice that the difference between these two roots
are 32/8 =4 which is as given in the problem
Exercise : Verify that p=5/2 also gives the
same result
2.19 Summary of
learning
No 
Points to remember 
1 
The
roots of quadratic equation ax^{2} +bx+ c = 0 are x = [b+_{}]/2a AND [b_{}]/2a 
2 
If m and n are roots of a
quadratic equation then the sum of the roots (m+n) = b/a 
3 
If m and n are roots of a
quadratic equation then the product of roots (mn) = c/a 
4 
If
m and n are roots of a quadratic equation then the equation is x^{2 }–(n+m)x
+mn =0 
2.19 Additional Points:
2.19.4 Binomial
theorem:
We have learnt that any algebraic expression with 2
variables is called a binomial. We also know that
(x+y)^{0}=1
(x+y)^{1}=x+y
(x+y)^{2}=x^{2}+2xy+y^{2}
(x+y)^{3}= x^{3}+3x^{2}y+3xy^{2}+y^{3}
Similarly
(x+y)^{4}= x^{4}+4x^{3}y+6x^{2}y^{2}+4xy^{3}+y^{4}
What are the observations?
1. The exponent of the first term(x) starts with
the exponent of the binomial (n) and in subsequent terms it decreases by 1 till
it is 0.
2. The exponent of the second term(y) starts with 0
and in subsequent terms it increases by 1 till it becomes equal to the exponent
of the
binomial.
3. The sum of exponents of x and y in each term is
equal to the exponent of the binomial.
4. There coefficients of first and last term is
always 1.
4. There is also a pattern among coefficients of other
terms as shown below.
The above triangle has come to be known as Pascal Triangle named after Pascal (16^{th}
Century AD). However this arrangement called as ‘Meru Prastara’ was known to Indian Mathematicians
much earlier and was first provided by Pingala (3^{rd }century BC).
Since this method of finding coefficients for
large values of n is difficult, we have the following theorem called the Binomial
theorem.
(x+y)^{n} = _{n}C_{0}x^{n}+_{
n}C_{1}x^{n1}y+_{ n}C_{2}x^{n2}y^{2}+………+_{
n}C_{r}x^{nr}y^{r}+……..+_{n}C_{n}y^{n}
Where the coefficient _{n}C_{r
} is defined as _{n}C_{r}=
n!/[(nr)!r!)] (Refer to section 1.9)
2.19.4 Problem 1: Find the 4^{th} term of [3a+(1/2a)]^{7}
By binomial theorem the 4^{th} term is T_{4 }= _{7}C_{3}_{ }x^{73}y^{3}= 7!/[4!3!)](3a)^{7}^{3}(1/2a)^{3}^{}
=(7*6*5*4!)/ [4!3!)]3^{4}a^{4}/(2^{3}a^{3})
= (35*81*a)/8
= (2835a/8)