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 or 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
x = [b
+_{}]/2a CxÀªÁ 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}+………+_{
nCrxnryr}+……..+_{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)