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 = x2
Thus x2 = 25 =5*5
x=5 Meters
Since 25 = -5*-5, x= -5 also satisfies the condition x2
= 25. We can say that x = 
5
are roots of the equation x2 = 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 x2 = 25 can also be expressed as x2 - 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 ax2
+c = 0 where a and c are real numbers
and a 
0, is called a ‘pure quadratic equation’ One example is 3x2 -16=0
2. An equation of the type ax2 +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 3x2
-5x-16=0
Note : If ab =0 then either a=0 or b=0 or both a =0 and b=0
Example : Let us solve 3x2 -16=0
3x2 =16 (By
transposition)
x2 =16/3
x = 
= 
/
= (4/)
2.19
Problem 1 : Solve x2/2 – 3/4 = 29/4
Solution:
On transposition we have
x2/2 = 29/4+3/4 = (29+3)/4 = 32/4 =8
x2
=16
x = 4
2.19
Problem 2 : Solve (2m-5)2= 81
Solution:
(2m-5)2= 92
2m-5 = 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=(2m-5)2=(9)2=81= RHS
When m = - 2: LHS=(2m-5)2=(-4-5)2=(-4-5)2=(-9)2=81=RHS
2.19
Problem 3 : If c2= a2+b2
Solve for b. If a=8 and c=17 find the value of b
Solution:
Given c2= a2+b2
b2= c2-a2
b = 
(c2-a2)
Substituting given
values for a and c in the above equation we get
b = (c2-a2)
= (172-82)
= (289-64)
= (225)
= 15
Verification:
When a=8 and b=15 we have RHS= a2+b2=64+225
=289 = 172= c2=LHS
2.19
Problem 4 : The volume of a cylinder of radius ‘r’ and height ’h’ is given by the formula Volume V = 
r2h
1. Solve for r.
2. Find the radius of the cylinder if Volume =176 and
height =14
Solution:
Assume = 22/7
Since V =r2h
r2= 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= r2h= 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 6-p2=p
Solution:
This is equivalent to solving p2+p-6 = 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
p2+p-6
= p2+3p-2p -6
= p(p+3) -2(p+3) ---- take out the common factor (p+3)
= (p+3)(p-2)
Since p2+p-6 = 0
(p+3)(p-2) = 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 p-2 = 0
p= -3 or p =2 are roots
of the given equation
Verification:
Let us put p=2 in the given
equation
LHS = 22+2-6
=4+2-6 = 0 =RHS. Similarly verify for p = -3
2.19.1
Problem 2: Solve
6 y2+y -15 = 0.
Solution:
We need to express LHS in the form(ax+b)(cx+d)={ acx2 + 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 y2+y -15
= 6 y2+10y -9y -15
= 2y(3y+5)-3(3y+5) – take out the
common factor 3y+5
= (3y+5)(2y-3)
Since 6 y2+y -15 =0
(3y+5)(2y-3) =0
This is possible if 3y+5 = 0 or 2y-3 =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(m2+1)
Solution:
This is equivalent to 6m2-13m+6
=0
We need to express LHS in the form(ax+b)(cx+d)={ acx2 + 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
6m2-13m+6=0
= 6m2-9m -4m+6
= 3m(2m -3) -2(2m-3) ------à take out the common factor 2m-3
= (2m-3)(3m-2)
Since 6m2-13m+6 =0
(2m-3)(3m-2)=0
This is possible if 2m-3 = 0 or 3m-2 =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 =(8-26)/3
+6 = 0 =RHS. Similarly try for m= 3/2
2.19.1
Problem 4: Solve
y2-2y+2 =0
Solution:
We need to express this equation in the form(ax+b)(cx+d)={ acx2 +
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 2x2+3x+1 =0
|
No |
Step |
Explanation |
|
1 |
x2
+(3/2)x+ (1/2) =0 |
Divide
both sides of the given equation by 2 |
|
2 |
x2+(3/2)x=
-(1/2) |
By
transposing (1/2) to RHS |
|
If we
can use the identity (x+a)2
= x2+2ax+ a2 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 |
x2+(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= x2 +2(3/4)x
+ (3/4)2= [x+(3/4)]2 |
p2+2pq+q2
= (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)) = |
Square
root of step 6 |
|
8 |
x =
-(3/4) |
Simplification |
We shall use the above method for solving generic equation
ax2 +bx+ c =0
Formula for finding roots of the
quadratic equation
Let us find the roots of the
Quadratic equation whose general form is ax2 +bx+ c =0, where a, b and c are real numbers and a 0 and b 0.
|
No |
Step |
Explanation |
|
1 |
x2
+(bx/a)+ (c/a) =0
|
Divide
both sides by ‘a’ |
|
2 |
x2
+(bx/a) = -( c/a) |
By
transposing c/a to RHS |
|
3 |
x2 +(bx/a)
+ (b/2a)2 = -( c/a) + (b/2a)2 |
By
adding (b/2a)2 to both sides |
|
4 |
LHS= x2 +(bx/a) + (b/2a)2= [x+(b/2a)]2 |
p2+2pq+q2
= (p+q)2 with p=x and q= b/2a |
|
5 |
RHS = b2/4a2-
c/a= (b2-4ac)/ 4a2 |
By
having common denominator as 4a2 |
|
6 |
[x+(b/2a)]2
=(b2-4ac)/ 4a2 |
By Step
4 and 5 as LHS=RHS |
|
7 |
x+(b/2a)
= = |
Take
square root of the last step |
|
8 |
x = [-b |
Transpose b/2a to RHS |
Therefore roots of the equation ax2 +bx+
c =0 are:
x = [-b +(b2-4ac)]/2a AND x = [-b -(b2-4ac)]/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
4x2+8x+4 = 0
Solution:
Here we have a =4, b=8, c =4
b2-4ac = 64 – 4*4*4 = 0
(b2-4ac)
= (0)
= 0
There fore as per the formula
roots are
p = [-b +
]/2a
=(-8+0)/8 = - 1 or
p = [-b -]/2a =
(-8-0)/8 = - 1
Here the roots are same = - 1
Alternatively, note the given equation is equivalent to 4(x2+2x+1)
= 4(x+1)(x+1) which again suggests that roots are -1.
2.19.1
Problem 6: Solve p2+p-6 = 0(Repetition of problem
2.19.1.1 solved earlier)
Solution:
This equation is of the form ax2
+bx+ c =0
Here we have a =1, b=1, c =-6
b2-4ac = 1 – 4*1*(-6) = 25
= (25)
= 5
As per the formula, roots are
p = [-b +]/2a
=(-1+5)/2 = 2 or
p = [-b -]/2a =
(-1-5)/2 = -3
These are the roots we got earlier
2.19.1
Problem 7: Solve 6y2+y -15 = 0(Repetition
of problem 2.19.1.2 solved earlier) and then factorise.
Solution:
This equation is of the form ax2
+bx+ c =0
Here we have a=6, b=1, c= -15
b2-4ac = 1 – 4*6*(-15) = 361
(b2-4ac)
= (361)
= 19
As per the formula, roots are
y = [-b +]/2a
=(-1+19)/12 = 18/12= 3/2 or
y = [-b -]/2a =
(-1-19)/12 = -20/12
= -5/3
These are the roots we got earlier
Since 3/2 and -5/3 are roots of the given equation, (y-3/2)(y+5/3) are factors of the given equation
Note (y-3/2)(y+5/3) =
(2y-3)(3y+5)/6
6y2+y
-15 = (2y-3)(3y+5)
Exercise:
Solve example 2.19.1.3 using the formula
method
2.19.1
Problem 8: Solve
y2-2y+2 =0(Repetition of problem 2.19.1.4 which was not solved earlier)
Solution:
This equation is of the form ax2
+bx+ c =0
Here we have a=1, b=-2, c=2
b2-4ac = 4 – 4*1*2 = -4
(b2-4ac)
= (-4)
= 2
As per the formula, roots are
y = [-b +]/2a
=(2+2)/2 = 1+ or
y = [-b -]/2a = (2-2)/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
y2-2y+2 = (1+)2 -2(1+) +2 (Use the formula (a+b)2
=a2+b2+2ab to expand (1+)2)
= [1 +(-1) +2 ] +[-2 -2] +2
=
1-1 +2 -2 -2+2 = 0 = RHS.
Similarly you can verify for other root= 1-
2.19.1
Problem 9: Solve
2(3y-1)/(4y-3) = 5y/(y+2) -2
Solution:
RHS = [5y -2(y+2)]/(y+2) = (3y-4)/(y+2)
We need to solve 2(3y-1)/(4y-3)
= (3y-4)/(y+2)
On cross multiplication we get 2(3y-1)*(y+2) =
(3y-4)*(4y-3)
i, e 2(3y2+6y –y -2) =
12y2-9y -16y+12
6y2+10y -4 = 12y2-25y
+12(By simplifying after transposing all terms from LHS to RHS we get:)
0 = 6y2-35y +16: 6y2-35y
+16=0
This equation is of the form ax2 +bx+ c =0
Here we have a=6, b=-35, c= 16
b2-4ac = 1225 – 4*6*16 = 1225-384 = 841
(b2-4ac)
= (841)
= 29
As per the formula, roots are
y = [-b +]/2a
=(35+29)/12 = 16/3 or \
y = [-b -]/2a =
(35-29)/12 = 1/2
Verification:
Substituting these values in the equation it can be
seen that LHS=RHS
2.19.1
Problem 10: Solve
(y-1)(5y+6) /(y-3) = (y-4)(5y+6)/(y-2)
Solution:
On cross multiplication in the
equation we get
(y-1)(5y+6)(y-2)
= (y-4)(5y+6)(y-3) on expanding terms on both LHS and RHS we get
LHS = (5 y2+6y-5y-6)(y-2)
= (5 y2+y-6)(y-2)
= 5 y3+ y2-6y -10 y2-2y+12
=5 y3 -9y2-8y+12
RHS
=
(5y2+6y-20y-24)(y-3)
= (5y2-14y -24)(y-3)
= 5y3-14 y2-24y -15y2+42y+72
= 5y3-29y2+18y+72
Since it is given that LHS=RHS we have
5 y3 -9y2-8y+12= 5y3-29y2+18y+72.
(On transposing all the terms from RHS to LHS we get:)
5 y3 -9y2-8y+12-(5y3-29y2+18y+72)
=0(On simplification we get)
20y2-26y-60 = 0 ( By
taking out 2 as a common factor)
10y2-13y-30 = 0
This equation is of the form ax2 +bx+ c =0
Here we have a=10, b=-13, c= -30
b2-4ac = 169 – 4*10*(-30) = 169+1200 = 1369
(b2-4ac)
= (1369)
= 37
As per the formula, roots are
y = [-b +]/2a
=(13+37)/20 =
50/20 = 5/2 or
y = [-b -]/2a =
(13-37)/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
[(y-1)/(y-3)] =[(y-4)/(y-2)]
: By cross multiplication we get
(y-1)(y-2) = (y-4)(y-3)
i,e
y2-2y-y+2 = y2-3y-4y+12
i,e
y2-3y+2 = y2-7y+12: (On transposition we get)
i,e
y2-3y+2-( y2-7y+12)=0
i,e
y2-3y+2-y2+7y-12=0
i,e
4y-10=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)]
= (y2+y2+2y+1)/(
y2+y)
Since LHS = RHS we get
(y2+y2+2y+1)/(
y2+y) = 25/12 (On cross multiplication we get)
12(y2+y2+2y+1) = 25( y2+y)
24y2+24y+12 = 25y2+25y.
On transposing LHS to RHS we get
0 = y2+y-12
This equation is of the form ax2 +bx+ c =0
Here
we have a=1, b=1, c= -12
b2-4ac = 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 = (-1-7)/2 = - 4
Verification:
Substituting these values in the
equation it can be seen that LHS=RHS
2.19.1
Problem 12 : Solve (3x2-5x+2) (3x2-5x-2)=21
Solution:
1. Let 3x2-5x
= y then solve for y in (y+2) (y-2) =21
2. Substituting
value for y in the equation 3x2-5x = 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
(x-8) people went for the picnic
The revised food bill per head =
480/(x-8)
This is given to be Rs 10 more
than what was planned earlier
The new rate = old rate +10
So we have 480/(x-8) = 480/x +
10
After simplifying RHS we get
480/(x-8) = (480+10x)/x. (On cross
multiplication we get)
480x = (480+10x)(x-8)
(On expanding RHS we get )
RHS= 480x -480*8 +10x*x-80x
= 480x - 3840+ 10x2-80x = 10x2+400x-3840
0 =10x2+400x-3840-480x. (By transposing 480x to RHS)
I.e. 0 =10x2-80x-3840. Dividing both the sides of this
equation by 10 we get
x2-8x-384 =0
This equation is of the form ax2 +bx+ c =0
Here we have a=1, b= -8, c= -384
b2-4ac = 64 – 4*1*(-384) = 64 +1536 =1600
= (160000)
= 40
As per the formula, roots are
x = [
] =(8+40)/2 = 24
or
x = [-b -]/2a = (8-40)/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 = x2+ y2
.It is given that hypotenuse =20 Since we are given that x-y = 4: We have x= 4+y.
Substituting this value of x in equation (1) and then expanding we have 400 =
x2+ y2 =(4+y)2+ y2
= (16+8y+ y2)+ y2=16+8y+ 2y2 . On transposing terms from LHS to RHS we have 0 = 2y2+8y-384.This
equation is of the form ax2 +bx+ c =0 Here we
have a=2, b= 8, c= -384
y = [-b - 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 = 122+ 162 = 144+ 256 = 400 =202
.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/x-1200/(x+30) = 2
Exercise : Apply the formula to get the
correct answer x=120
Verification:
1200/120 – 1200/150 = 10-8 =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/x-2(Speed reduces by the speed of stream)
Total
time taken = 8/(x-2) + 8/(x+2) which is given to be 5/3
Thus the equation to be solved is
8/(x-2) + 8/(x+2) = 5/3
Exercise: Apply the formula to get the correct answer x =10
Verification:
Total time taken = 8/(10-2) + 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) = (3000-2x)*(x+250)/2x
I.e. 3000x = (3000-x)(x+250) ( By cross multiplication)
I.e. 3000x = 3000x -x2+750000-250x
I.e. x2-750000+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
(x-20)
= 8
x2-40x+400
= 64x --------à (a+b)2Formula).
x2-104x+400
=0
(x-100)*(x-4)
=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 1/5th 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
(x2/25) –(6x/5)+9+1=x
(x2/25) –(11x/5)+10=0
x2–55x+250=0
(x-50)*(x-5) =0
x=50 Or x=5. It can not be 5, as (x/5)-3 can not be negative.
Verification:
50= (10-3)2+1= 49+1,
2.19.1 Problem 21: Solve 12( x2+
1/ x2) -56(x+1/x) = -89
Solution:
The given equation is 12( x2+ 1/ x2) -56(x+1/x) +89 =0
We know (x+1/x)2
= x2+ 1/ x2+2
x2+
1/ x2=(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 12y2-56y+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. (x2+1)/x = 5/2
I.e. 2(x2+1) = 5x
I.e. 2x2-5x+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. (x2+1)/x = 13/6
I.e. 6(x2+1) = 13x
I.e. 6x2-13x+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 b2-4ac in
solving the problems?
In problem 2.19.1 .5, we have seen
that b2-4ac = 0 and roots are same
In problem 2.19.1 .8, we have seen that b2-4ac
<0 and roots are not real numbers
In all other examples we have seen
that b2-4ac > 0 and roots are real numbers
The expression b2-4ac
is called ‘discriminant’
and is denoted by 
(called delta)
We conclude the following:
|
|
Value of
Discriminant(b2-4ac) = |
Nature of
roots=[-b |
|
1 |
|
Roots are real and equal |
|
2 |
|
Roots are real and distinct(not equal) |
|
3 |
|
Roots are imaginary(not real) and distinct |
2.19.2 Problem 1: For what positive values of ‘m’, the roots of mk2-3k+1
=0 are equal, (real and distinct) and (imaginary and distinct)
Solution:
Here we have a=m, b= -3, c= 1
b2-4ac = 9 – 4m
1. Roots are equal when b2-4ac =0
(I.e. 9-4m =0, i.e. m = 9/4)
2. Roots are real and not equal when b2-4ac
>0
(I.e. 9-4m >0, i.e. 9 >4m, i.e. m < 9/4)
3. Roots are imaginary and not equal when b2-4ac
<0
(I.e. 9-4m <0, i.e. 9 <4m, i.e. m > 9/4)
2.19.2 Problem 2: For what values of ‘m’, the roots of r2-(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
b2-4ac = (m+1) 2-16
= [(m+1)+4]*[(m+1)-4] ===> By factorization
= (m+5)(m-3)
1. Roots are equal when b2-4ac =0
(i.e. (m+5)= 0 or (m-3)=0 i.e. m=-5 or m=3
2. Roots are real and not equal when b2-4ac
>0
(i.e. (m+5)(m-3) >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 m-3>0
I.e. m> -5 and m>3: This is possible only if
m>3
Case 2: both m+5
< 0 and m-3 <0
I.e. m< -5 and m<3: This is possible only if m
<-5
3. Roots are imaginary and not equal when b2-4ac
<0
(i.e. (m+5)(m-3) <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 m-3>0
I.e. m< -5 and m>3: This is not possible at all
Case 2: m+5 > 0 and m-3 <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) n2+2(p+3)n +(p+8) =0
has equal roots
Solution:
This equation is of the form ax2
+bx+ c =0
Here we have a=(p+1), b= 2p+6, c=
p+8
b2-4ac = (2p+6)2 – 4*(p+1)(p+8) = (4p2+24p+36) -4(p2+8p+p+8)=
4p2+24p+36 -4p2-36p-32 =-12p+4
If the roots are to be equal then b2-4ac =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) n2+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+25-16p-28)/4
= (9p-3)/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) c2+2(p+1)c +p =0 has equal roots
Solution:
This equation is of the form ax2
+bx+ c =0
Here we have a=(3p+1), b= 2p+2,
c= p
b2-4ac = (2p+2)2 – 4*(3p+1)p = (4p2+4+8p) -4(3p2+p)=
4p2+4+8p -12p2-4p = -8p2+4p+4 = - 4(2p2-p-1)
If the roots are to be equal then b2-4ac =0
I.e. 2p2-p-1 = 0
LHS = 2p2-2p+p-1 = 2p(p-1)+(p-1)
= (p-1)(2p+1)
Since 2p2-p-1 = 0 we have
(p-1)(2p+1) = 0
p=1 or p= -1/2 are the answers
NOTE: To find roots of 2p2-p-1 = 0 we used
factorisation method.
As per the formula, roots with p=1 is
c = [-b +]/2a
=[-2p-2
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) c2+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) c2+2(p+1)c
+p =0
2.19.2 Problem 5: Find value of ‘p’ for which 2y2-py +1 =0
has equal roots
Solution:
This equation is of the form ax2
+bx+ c =0
Here we have a=2, b= -p, c= 1
b2-4ac = p2 -8
If the roots are to be equal then b2-4ac =0
I.e.
p2 = 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
co-efficients:
Let ‘m’
and ‘n’ be the roots of quadratic equation of the form ax2
+bx+ c =0
(x-m)(x-n) = 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)(a-b)
we get
= [ (-b)2- {}2] /4a2
= [b2 -(b2-4ac) ] /4a2
= 4ac/4a2
= 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 x2 +(ab)x+ (a+b) =0
Solution:
This equation is of the form ax2
+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 pr2 = r-5
Solution:
This is equivalent to pr2
–r+5= 0
This equation is of the form a x2 +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 ax2
+bx+ c =0
Then we know (x-m)(x-n) =
0
But (x-m)(x-n)
=x(x-n)-m(x-n)
= x2 –xn –mx +mn
= x2 –x(n+m) +mn
= x2 –(n+m)x +mn
The general format is
x2
–(sum of roots)x +(product of roots) =0
2.19.4
Problem 1: If ‘p’
and ‘q’ are the roots of the equation 2a2-4a+1=0 find the value of (p+q)2+4pq and p3 +q3 and
also form the equation whose roots are p3 and q3
Solution:
This equation is of the form ax2+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 a3+b3=
(a+b) (a2+b2-ab)
p3
+q3
= (p+q)(
p2+q2-pq)
= (p+q)[( p2+q2+2pq) -3pq)]
= (p+q)[( p+q)2-3pq]
=2*[4-3/2] =5 (By
substituting vales for (p+q) and pq )
We are also required to form an equation whose roots are p3
and q3
Sum of roots = p3 +q3 =5( we had just calculated above)
Product of roots = p3*q3 = (pq)3
=(1/2)3 =1/8
The desired equation is
x2-(sum of roots)x+ (product of
roots)= 0
I.e. x2-5x+ 1/8= 0 (by multiplying terms by 8
we get)
I.e. 8x2-40x+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 = (p2+q2)/pq:
mn = p/q*q/p =1
The
standard form is x2 –(n+m)x +mn= 0
I.e. x2 –(p2+q2)x/pq +1 = 0
I.e. (pqx2 –(p2+q2)x
+pq)/pq =0( Have pq as common denominator)
I.e. pqx2 –(p2+q2)x +pq=0
2.19.4
Problem 3: If one
root of the equation x2+px+q=0 is 3
times the other prove that 3p2=16q
Solution:
This equation is of the form ax2+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 = 3n2
3p2= 3(-4n)2= 48 n2=16*3n2 = 16q(
3n2=q)
2.19.4
Problem 4: Find
the value of ‘p’ so that the equation 4x2-8px+9=0 has roots whose difference is 4
Solution:
This equation is of the form ax2+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= 2p-4: i,e m=p-2 ------à(3)
By substituting n=
m+4 in (2) we get
m(m+4) =9/4
I.e. m2+4m - 9/4 =0
I.e. (p-2)2+4(p-2) - 9/4 =0(m=p-2 as per (3))
I.e. p2-4p+4 +4(p-2) - 9/4 =0 (By expanding
(p-2)2 using formula )
I.e. p2-4p+4 +4p-8 - 9/4 =0
I.e. p2-4 - 9/4 =0
I.e. p2-25/4 =0
I.e. p2= 25/4
p = 5/2
Verification: Substitute value of p (-5/2) in the given equation we get
4x2-8px+9=0
i.e. 4x2-8*(-5/2)x+9=0
i.e. 4x2+20x+9=0 This is of the form
ax2+bx+c=0 where a=4, b=20,
c=9
b2-4ac = 400 – 4*4*9 = 400-144 =256
= (256)
= 16
As per the formula, roots are
x = [-b +]/2a
=(-20+16)/8 =
-4/8
x = [-b -]/2a =
(-20-16)/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 ax2 +bx+ c =
0 are x = [-b+ [-b- |
|
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 x2 –(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=x2+2xy+y2
(x+y)3=
x3+3x2y+3xy2+y3
Similarly
(x+y)4=
x4+4x3y+6x2y2+4xy3+y4
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 co-efficients of first and last term is always 1.
4. There is also a pattern among co-efficients of other terms
as shown below.
The above triangle has come to be known as Pascal Triangle named after Pascal (16th
Century AD). However this arrangement called as ‘Meru Prastara’ was known to Indian Mathematicians much
earlier and was first provided by Pingala (3rd century BC).
Since this method of finding co-efficients for large values
of n is difficult, we have the following theorem called the Binomial theorem.
(x+y)n
= nC0xn+ nC1xn-1y+
nC2xn-2y2+………+ nCrxn-ryr+……..+nCnyn
Where the co-efficient nCr is defined as nCr=
n!/[(n-r)!r!)]
(Refer to section
1.9)
2.19.4
Problem 1: Find
the 4th term of [3a+(1/2a)]7
By binomial theorem the 4th term is T4 =
7C3
x7-3y3= 7!/[4!3!)](3a)7-3(1/2a)3
=(7*6*5*4!)/ [4!3!)]34a4/(23a3)
= (35*81*a)/8
= (2835a/8)