new suppor quadralics
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3. Solving quadratic equation by formula method
General form of a quadratic equation ax2 + bx+ c = 0
Divide by a 0a
c
a
bx
a
ax2
=++
Transpose the constant term to R.H.S. a
c
a
bx
x
2
=+
Add
2
a2
b
to both the sides
22
2
a2
b
a
c
a2
b
a
bxx
+=
++
2
22
a4
b
a
c
a2
bx +=
+
2
22
a4
bac4
a2
bx +=
+
Simplify 2
22
a4
ac4b
a2
bx
=
+
Taking square root 2
2
a4
ac4b
a2
bx
=+
a2
ac4b
a2
bx
2 =+
a2
ac4b
a2
bx
2
=
Roots are a2ac4bb
x
2 =
a2
ac4bbx
2 += or
a2
ac4bbx
2 =
Roots of the equation ax2 + bx+ c = 0 are x =a2
ac4bb 2
Note : The roots of the equation ax2 + bx+ c = 0 can also be found using
Sridharas method.
x x
x
x
x
x
x
x
x
x
x x
x
x
x
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Example 1 : Solve the equation x2 7x+ 12 = 0
consider x2 7x+ 12 = 0
This is in the form ax2 + bx+ c = 0
the coefficients are a = 1, b = 7 & c = 12
The roots are given by x = a2
ac4bb 2
Substituting the values a = 1, b = 7 and c = 12
1x2
)12)(1(4)7()7(x
2 =
2
48497x
=
Simplify x =2
17
x =2
17 +or x =
2
17
x =2
8or x =
2
6
Roots are x = 4 or x = 3
Example 2 : Solve the equation 2p2 p = 15
Consider 2p2 p = 15
2p2 p 15 = 0
This is in the form ax2 + bx+ c = 0
The coefficients are a = 2, b = 1 and c = 15
The roots are given bya2
ac4bbx
2 =
Substituting the values a = 2, b = 1 and c = 15
p =)2(2
)15)(2(4)1()1(2
x
x
x
x
x x
x
x x
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x
p =4
12011 ++
p =4
1211
p =4
111
p =4
111+or p =
4
111
p =4
12or p =
4
10
p = 3 or p =2
5
Example 3 : Solve the equation 2k2 2k 5 = 0
Consider 2k 2 2k 5 = 0
This is in the form ax2 + bx+ c = 0
The coefficients are a = 2, b = 2 and c = 5
The roots are given by x =a2
ac4bb 2
Substituting the values a = 2, b = 2 and c = 5
k =
)2(2
)5)(2(4)2()2( 2
k =4
4042 + =
4
442
k =4
1122 =
4
1112
The roots are k = 2111+
or k = 2
111
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Example 4 : Solve the equation m2 2m = 2
Consider m2 2m = 2
m2 2m 2 = 0
This is in the form ax2 + bx+ c = 0
Comparing the coefficients a = 1, b = 2 and c = 2
The roots are given by x =a2
ac4bb2
m =)1(2
)2)(1(4)2()2( 2
m =2
842 ++
m =2
122
m =2
312
m = 31+ or m = 31Exercise : 5.3
Solve the following equations by using formula
1) a2 2a 4 = 0 2) x2 8x+ 1 = 0 3) m2 2m + 2 = 0
4) k2 6k = 1 5) 2y2 + 6y = 3 6) 8r2 = r + 2 7) p = 5 2p2
8) 2z2 + 7z + 4 = 0 9) 3b2 + 2b = 2 10) a2 = 4a + 6
4. Equations reducible to the form ax2 + bx + c = 0
Example 1 : Solve the equation (x+ 6) (x+ 2 ) =x
Solution :(x
+ 6) (x
+ 2 ) =x
x2 + 6x+ 2x+ 1 2 =x
x2 + 8x+ 1 2 x= 0
x2 + 7x+ 12 = 0
x2 + 4x+ 3x+ 12 = 0
x(x+ 4) + 3 (x+ 4) = 0
(x+ 4) (x+ 3) = 0
Either (x+ 4) = 0 or (x+ 3 ) = 0
x
= 4 orx
= 3
x
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Example 2 : Solve the equation (a 3)2 + (a + 1)2 = 16
Solution : (a 3)2 + ( a + 1 )2 = 16
Using (a + b)2 = a2 + 2 a b + b2
(a b)2 = a 2 a b + b2
[(a)2 2(a)(3) + 32] + [a2 + 2 (a) (1) + 12] = 16
a2
6a + 9 + a2
+ 2a + 1 16 = 02a2 4a 6 = 0
a2 2a 3 = 0
a2 3a + 1a 3 = 0
a ( a 3 ) + 1 ( a 3 ) = 0
(a 3) (a + 1) = 0
Either (a 3) = 0 or (a + 1) = 0
a = 3 or a = 1
Example 3 : Solve 5(p 2)2 + 6 = 1 3 ( p 2 )
Solution : 5(p 2)2 + 6 = 1 3 ( p 2 )
Let p 2 = b
then 5b2 + 6 = 1 3 b
5b2 1 3 b + 6 = 0
5b2 1 0 b 3 b + 6 = 0
5b (b 2) 3 (b 2) = 0
(b 2) (5b 3) = 0
Either (b 2) = 0 or (5b 3) = 0
b = 2 or b = 5
3
p 2 = 2 or p 2 = 5
3(
b = p 2)
p = 2 + 2 or p = 5
3 +
1
2
p = 4 or p =5
13
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Example 4 : Solve the equation1k
1k
5k2
2k3
+
=++
Consider1k
1k
5k2
2k3
+
=++
Cross multiplying (3k + 2) (k 1) = (2k + 5) (k + 1)
3k2
+ 2k 3k 2 = 2k 2
+ 5k + 2k + 53k2 1 k 2 2 k 2 7k 5 = 0
3k2 1 k 2 2 k 2 7k 5 = 0
On simplification k2 8k 7 = 0
This is in form of ax2 + bx+ c = 0
The co-efficients are a = 1, b = 8, c = 7
The roots of the equationa2
ac4)b(bx
2 +=
k =1x2
)7)(1(4)8()8(2
k =2
28648 ++
k = 2
928
k =2
2328 =
( )2
2342
k = 234
Example 5 : Solve the equation 1
y2
3
4
y=
Consider 1y2
3
4
y=
Taking L.C.M. 1y4
6y2=
By cross multiplication y2 6 = 4y
y2 4y 6 = 0
x
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this is in the form ax2 + bx+ c = 0
comparing coefficients a = 1, b = 4, c = 6
y =)1(2
)6)(1(4)4()4( 2
the roots of the equation are =2
24164 +
y =2
404
y =2
1024 =
( )2
1022
y = 102+ or y = 102
Example 6 : Solve1m2
4
3m
1
2m
4
+=
+
+
1m2
4
)3m)(2m(
)2m(1)3m(4
+=
++++
1m2
4
6m3m2m
2m12m42 +=+++
+
1m2
4
6m5m
10m32 +
=++
+
On Cross multiplying, 4m2 + 20m + 24 = 6m2 + 20m + 3m + 10
4m2 + 20m + 24 6m2 2 3 m 1 0 = 0
2m2 3m + 14 = 0
This is in the Standard form 2m2 + 3m 14 = 0
2m2 + 7m 4m 14 = 0
m(2m + 7) 2 (2m + 7) = 0
(2m + 7) (m 2) = 0
Either (2m + 7) = 0 or (m 2) = 0
m =
2
7or m = 2
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Example 2 : Sum of a number and its reciprocal is 55
1. Find the number.
Solution : Let the number be = y
Reciprocal of the number =y
1
(Number) + (its reciprocal) = 55
1
y +y
1 =
5
26
y
1y2 + =
5
26
5(y2 + 1) = 26y
5y2 + 5 = 2 6 y
5y2 2 6 y + 5 = 0
5y2 2 5 y 1 y + 5 = 0
5y (y 5) 1 (y 5) = 0
(y 5) (5y 1) = 0
Either (y 5) = 0 or (5y 1) = 0
y = 5 or y = 5
1
The required number is 5 or 51
Example 3 :The base of a triangle is 4 cms longer than its altitude. If the area of
the traingle is 48 sq cms. Find the base and altitude. Solution : Let the altitude = x cms.
Base of the triangle = (x+ 4) cms.
Area of the triangle = 2
1 (base) (height)
48 = 2
1 (x+ 4)x
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[ ] [ ]
48 x 2 = (x+ 4)x
96 = x2 + 4x
x2 + 4x 96 = 0
x2 + 12x 8x 96 = 0
x(x+ 1 2 ) 8 (x+ 1 2 ) = 0
(x+ 12) (x 8) = 0
Either (x+ 12) = 0 or (x 8) = 0
x= 12 or x= 8
Altitude =x= 8 cms.Base = x+ 4
= 8 + 4
= 12 cms
Example 4 : Rashmi bought some books for Rs. 60. Had she bought 5 more books
to the same amount each book would have cost her 1 rupee less. Find
the number of books bought by Rashmi and price of each book.
Solution : Let the number of books = x
Total cost of the books = Rs. 60
Cost of each book = Rs. x
60
If number of books is (x+ 5)
Then the cost of each book = Rs.)50(
60
+x
Difference in cost = 1 Re.
Cost of each book cost of each book if number of books if number of books = Difference amount
is (x) is (x+5)
x
60
5
60
+x = 1
)5(
60)5(60
++
xx
xx
= 1
x = 12 cannot be
considered, because
the length is always
positive
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[ ] [ ]
xx
xx
5
60300602 +
+ = 1
1
1
5
3002
=+ xx
x
2 + 5x= 300
x2 + 5x 300 = 0
x2 + 20x 15x 3 0 0 = 0
x(x+ 2 0 ) 1 5 (x+ 2 0 ) = 0
(x+ 20) (x 1 5 ) = 0
Either (x+ 20) = 0 or (x 1 5 ) = 0
x
= 20 orx
= 15 Number of books = x = 15
Cost of each book =x
60 =
15
60 = Rs. 4
Example 5 : The speed of a motor boat in still water is 15 km/hr. If it goes down
the stream 30 kms and again returns to the starting point in total time
of 4 hrs and 30 minutes, find the speed of the stream.
Solution : Speed in Still water is = 15 km/hr
Total distance travelled = 30 km
Let the speed of the stream = x km/hr
Speed up the stream = (15 x) km/hr
Speed down the stream = (15 + x) km/hr
Total time taken = 4hrs and 30 minutes
Time taken to row down the stream =x+15
30
Time taken to row up the stream =x15
30
Time taken to row + time taken to row = 4 hours 30 minutes
Down the stream up the stream
x+15
30+
x15
30=
2
14
x = 20 cannot be
considered because
number of books is always
positive
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2
9
)15()15(
)15(30)15(30=
+++xx
xx
2
9
225
30450304502
=
++x
xx
2
9
225
9002=
x
2225
9
)2(900x=
200 = 225 x2
x2 = 225 200
x2 = 25
x = 5
x = + 5 or x = 5
Speed of the stream = x = 5km/hr
Exercise : 5.5
1) The sum of a number and twice its square is 105. Find the number.
2) Product of two consecutive integers is 182. Find the integers.
3) The sum of the squares of three consecutive natural numbers is 194. Find the integers.
4) The length of rectangular field is 3 times its breadth. If the area of the field is
147 sq mts. Find the length of the field.
5) Hypotenuse of a right-angled triangle is 20 mts. If the difference between the lengths
of other two sides is 4 mts. Find the measures of the sides.
6) An Aero-plane takes 1 hr. less for a journey of 1200 km. If its speed is increasedby 60 km/hr from its initial speed find the initial speed of the plane.
7) Some students planned a picnic. The budget for the food was Rs. 480. As eight
of them failed to join the party the cost of the food for each member increased
by Rs. 10. How many students participated in the picnic?
8) Sailor Raju covered a distance of 8 km in 1 hr 40 minutes down stream and returns
to the starting point. If the speed of the stream is 2 km/hr, find the speed of the
boat in still water.
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9) A dealer sells an article for Rs. 24 and gains as much percent as the cost price
of the article. Find the Cost price of the article.
10) Sowmya takes 6 days less than the number of days taken by Bhagya to complete
a piece of work. If both Sowmya and Bhagya together can complete the same
work in 4 days. In how many days will Bhagya complete the work?
6. Nature of the roots of a quadratic equation.1) Consider the equation x2 2x+ 1 = 0
This is in the form ax2 + bx+ c = 0
The coefficients are a = 1, b = 2, c = 1
x =a2
ac4bb 2
x =1x2
1x1.4)2()2( 2 +
x =2
442
x =2
02+
x =2
02+ or x =2
02
x = 1 or x = 1 roots are equal
2) Consider the equation x2 2x 3 = 0
This is in the form ax2 + bx+ c = 0
the coefficients are a = 1, b = 2, c = 3
x =a2
ac4bb 2
x =1x2
16)2(
x =2
42+
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x =2
42+or x =
2
42
x =2
6or x =
2
2
x = 3 or x = 1 roots are distinct
3) Consider the equation x2 2x+ 3 = 0
This is in the form ax2 + bx+ c = 0
The coefficients are a = 1, b = 2, c = 3
x =a2
ac4bb2
x =1x2
)3)(1(4)2()2(22
x =2
1242
x =2
82
x =2
222
x =( )
2
212 = 21
x = 21 + or 21 roots are imaginary
From the above examples it is clear that,
1) Nature of the roots of quadratic equation depends upon the value of (b2 4ac)
2) The Expression (b2 4ac) is denoted by (delta) which determines the natureof the roots.
3) In the equation ax2 + bx+ c = 0 the expression (b2 4ac) is called the discriminant.
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Discriminant (b2 4ac) Nature of the roots
= 0 Roots are real and equal > 0 (Positive) Roots are real and distinct < 0 (negative) Roots are imaginary
Example 1 : Determine the nature of the roots of the equation 2x2 5x 1 = 0.
Consider the equation 2x2 5x 1 = 0
This is in form of ax2 + bx+ c = 0
The co-efficient are a = 2, b = 5, c = 1
= b2 4ac = (5)2 4(2) (1) = 25 + 8 = 33
> 0Roots are real and distinct
Example 2 : Determine the nature of the roots of the equation 4x2 4x+ 1 = 0
Consider the equation 4x2 4x+ 1 = 0
This is in the form of ax2 + bx+ c = 0
The co-efficient are a = 4, b = 4, c = 1
= b2 4ac
= (4)2 4 (4) (1)
= 16 16 = 0 Roots are real and equal
Example 3 : For what values of m roots of the equation x2 + mx+ 4 = 0 are
(i) equal (ii) distinct
Consider the equation x2 + mx+ 4 = 0
This is in the form ax2 + bx+ c = 0
the co-efficients are a = 1, b = m, c = 4
= b2 4ac = m2 4(1) (4) = m2 16
1) If roots are equal = 0 m2 16 = 0
m2 = 16
m = 16 m = 4
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2) If roots are distinct > 0 m2 16 > 0 m2 > 16
m2 > 16
m > 4
Example 4 : Determine the value of k for which the equation kx2 + 6x+ 1 = 0 has
equal roots.
Consider the equation kx2 + 6x+ 1 = 0
This is in the form ax2 + bx+ c = 0
the co-efficients are a = k, b = 6, c = 1
= b2 4ac
since the roots are equal, b2
4ac = 0 (
= 0)(6)2 4(k)(1) = 0
3 6 4 k = 0
4k = 36
k =4
36 = 9
k = 9
Example 5 : Find the value of p for which the equation x2 ( p + 2 )x+ 4 = 0 hasequal roots.
Consider the equation x2 ( p + 2 )x+ 4 = 0
This is in the form ax2 + bx+ c = 0
Coefficients are a = 1, b = (p + 2), c = 4
since the roots are equal = 0b2 4 a c = 0
[(p + 2)]2 4(1)(4) = 0
(p + 2)2 16 = 0
p + 2 = 16p + 2 = 4p + 2 = + 4 or p + 2 = 4
p = 4 2 or p = 4 2
p = 2 or p = 6
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If m and n are the roots of the
quadratic equation
ax2 + bx+ c = 0
Sum of the rootsa
b=
Product of rootsa
c+=
Exercise : 5.6
A. Discuss the nature of roots of the following equations
1) y2 7y + 2 = 0 2) x2 2x+ 3 = 0 3) 2n2 + 5n 1 = 0
4) a2 + 4a + 4 = 0 5) x2 + 3x 4 = 0 6) 3d2 2d + 1 = 0
B. For what positive values of m roots of the following equations are
1) equal 2) distinct 3) imaginary
1) a2 ma + 1 = 0 2) x2 mx + 9 = 0
3) r2 (m + 1) r + 4 = 0 4) mk 2 3k + 1 = 0
C. Find the value of p for which the quadratic equations have equal roots.
1) x2 px+ 9 = 0 2) 2a2 + 3a + p = 0 3) pk 2 1 2 k + 9 = 0
4) 2y2 py + 1 = 0 5) (p + 1) n2 + 2 ( p + 3 ) n + ( p + 8 ) = 0
6) (3p + 1)c2
+ 2 (p + 1) c + p = 0
7. Relationship between the roots and co-efficient of the terms of the quadratic
equation.
If m and n are the roots of the quadratic equation ax2 + bx+ c = 0 then
m =a2
ac4bb 2 +, n =
a2
ac4bb 2
m + n =a2
ac4bb 2 + +a2
ac4bb 2
m + n =a2
ac4bbac4bb 22 +
m + n =a2
b2
m + n =a
b-
mn =
+a2
ac4bb2
a2
ac4bb2
mn =
( )2
222
a4
ac4b)b(
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mn =( )
2
22
a4
ac4bb
mn = 2
22
a4
ac4bb +
mn = 2a4ac4 = ac mn = ac
Example 1 : Find the sum and product of the roots of equation x2 + 2x+ 1 = 0
x2 + 2x+ 1 = 0
This is in the form ax2 + bx+ c = 0
The coefficients are a = 1, b = 2, c = 1
Let the roots be m and n
i) Sum of the roots m + n =a
b =
1
2
m + n = 2
ii) Product of the roots mn =a
c =
1
1
mn = 1
Example 2 : Find the sum and product of the roots of equation 3x2 + 5 = 0
3x2 + 0x+ 5 = 0
This is in the form ax2 + bx+ c = 0
The coefficients are a = 3, b = 0, c = 5
Let the roots are p and q
i) Sum of the roots p + q =a
b =
3
0
p + q = 0
ii) Product of the roots pq =a
c =
3
5 pq =
3
5
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If m and n are the roots then the Standard form of the equation is
x2 (Sum of the roots)x + Product of the roots = 0
x2 (m + n) x + mn = 0
Example 1 : Form the quadratic equation whose roots are 2 and 3
Let m and n are the rootsm = 2, n = 3Sum of the roots = m + n = 2 + 3
m + n = 5Product of the roots = mn
= (2) (3)
mn = 6Standard form x2 ( m + n )x+ mn = 0
x2 (5)x+ ( 6 ) = 0
x2 5x+ 6 = 0
Example 2 : Form the quadratic equation whose roots are5
2 and
2
5
Let m and n are the roots
m =5
2 and n =
2
5
Sum of the roots = m + n =5
2 +
2
5 =
10
254+
m + n =10
29
Product of the roots = mn =2
5x
5
2 mn = 1
Standard form x2 (m + n)x+ mn = 0
x2 10
29x+ 1 = 0
10x2 29x+ 10 = 0
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Example 3 : Form the quadratic equation whose roots are 3 + 2 5 and 3 2 5
Let m and n are the roots
m = 3 + 2 5 a n d n = 3 2 5Sum of the roots = m + n
= 3 + 2 5 + 3 2 5
m + n = 6Product of the roots = mn
= (3 + 2 5 ) (3 2 5 )
= (3)2 (2 5 )2
= 9 20
mn = 11x
2 ( m + n )x+ mn = 0
x2 6x 11 = 0
Example 4 : If m and n are the roots of equation x2 3x+ 1 = 0 find the value
of (i) m2n + m n2 (ii)n
1
m
1+
Consider the equation x2 3x+ 1 = 0
This is in the form ax2 + bx+ c = 0
The coefficients are a = 1, b = 3, c = 1Let m and n are the roots
i) Sum of the roots m + n =a
b =
1
)3( = 3
m + n = 3
ii) Product of the roots mn =a
c
mn =1
1 mn = 1
(i) m2n+mn2 = mn (m + n)
= 1(3) = 3
(ii)m
1 +
n
1=
mn
mn + =
mn
nm + =
1
3
m1
+ n
1 = 3
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Example 5 : If m and n are the roots of equation x2 3x + 4 = 0 form the
equation whose roots are m2 and n2.
Consider the equation x2 3x+ 4 = 0
The coefficients are a = 1, b = 3, c = 4
Let m and n are the roots
i) Sum of the roots = m + n =a
b =
1
)3(
m + n = 3
ii) Product of the roots = mn =a
c =
1
4
mn = 4
If the roots are m2 and n2Sum of the roots m2 + n2 =(m+n)2 2mn
= (3)2 2(4)
= 9 8
m2 + n2 = 1
Product of the roots m2n2 = (mn)2
= 42
m2n2 = 16x
2 (m2 + n2)x+ m2n2 = 0
x2 (1)x+ (16) = 0
x2 x+ 16 = 0
Example 6 : If one root of the equation x2 6x+ q = 0 is twice the other, find the
value of q
Consider the equation x2 6x+ q = 0
This is in the form ax2 + bx+ c = 0
The coefficients are a = 1, b = 6, c = q
Let the m and n are the roots
i) Sum of the roots m + n =a
b =
1
)6(
m + n = 6
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Exercise : 5.8
A. Form the equation whose roots are
1) 3 and 5 2) 6 and 5 3) 2 and2
34)
3
2 and
2
3
5) 2 + 3 and 2 3 6) 3 + 2 5 and 3 2 5
B.
1) If m and n are the roots of the equation x2 6x+ 2 = 0 find the value of
i) (m + n) mn ii)m
1 +
n
1
2) If a and b are the roots of the equation 3m2 = 6m + 5 find the value of
i)ab
ba + ii) (a + 2b) (2a + b)
3) If p and q are the roots of the equation 2a2 4a + 1 = 0 Find the value of
i) (p + q)2 + 4pq ii) p3 + q3
4) Form a quadratic equation whose roots areq
p and
p
q
5) Find the value of k so that the equationx2 + 4x+ (k + 2) = 0 has one root equal
to zero.
6) Find the value of q so that the equation 2x2 3qx+ 5q = 0 has one root which
is twice the other.
7) Find the value of p so that the equation 4x2 8px+ 9 = 0 has roots whose
difference is 4.
8) If one root of the equationx2 + px+ q = 0 is 3 times the other prove that 3p2 = 16q
Graphical method of solving a Quadratic Equation
Let us solve the equation x2 4 = 0 graphically,
x2 4 = 0
x2 = 4let y =x2 = 4
y =x2
and y = 4
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y =x2
x= 0 y = 02 y = 0
x= 1 y = 12 y = 1
x= 2 y = 22 y = 4
x= 1 y = (1)2
y = 1x = 2 y = (2)2 y = 4
x 0 1 1 2 2 3
y 0 2 2 8 8 6
(x, y) (0, 0) (1, 2) (1, 2) (2, 8) (2, 8) ( 3 ,6)
Step 1: Form table of
corresponding values
of x and y
Satisfying the equation
y =x2
Step 2: Choose the scale onx axis, 1 cm = 1 unit
y axis, 1 cm = 1 unit.
Step 3: Plot the points (0, 0);
(1, 1); (1, 1); (2, 4)
and (2, 4) on graph
sheet.
Step 4: Join the points by asmooth curve.
Step 5: Draw the straight line
y = 4 Parallel to x-axis
Step 6: From the intersecting
points of the curve and
the line y = 4, draw
perpendiculars to thex axis
Step 7: Roots of the equations are x = +2 or x = 2
The graph of a quadratic polynomial is a curve called parabola
Example 1 : Draw a graph of y = 2x2 and find the value of 3 , using the graph.
Step 1: Form the table of
corresponding values ofx and y satisfying the
equation y = 2x2
Step 2: Choose the scale on x
axis, 1 cm = 1 unit and
y axis, 1 cm = 1 unit
Step 3: Plot the points (0, 0);
(1, 2) (1, 2); (2, 8) and
(2, 8) on graph sheet.
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x 0 1 1 2 2
y 0 1 1 4 4
(x, y) (0, 0) (1, 1) (1, 1) (2, 4) (2, 4)
x 0 1 1 2 2
y 2 1 3 0 4
(x, y) (0, 2) (1, 1) (1, 3) (2, 0) (2, 4)
Step 4: Join the points by a
smooth curve
Step 5: Draw the straight line
y = 6 Parallel tox-axis.
Step 6: From the intersecting
points of the curve and
the line y = 6, draw
perpendiculars to the
x-axis.
Step 7: Value of 3 = 1.7x = 1.7 or x = + 1.7
Example 2 : Draw a graph of y = x2 and y = 2-x and hence solve the equation
x2 +x 2 = 0
Step 1: Form the table of
corresponding values of
x and y satisfying the
equation y = x2
Step 2: Form the table of
corresponding values ofx and y satisfying the
equation y = 2 x.
Step 3: Choose the scale on x
axis 1 cm = 1 unit and
y axis, 1 cm = 1 unit.
Step 4: Plot the points (0, 0);
(1, 1); (1, 1); (2, 4)and (2, 4) on the graph
sheet.
Step 5: Join the points by a
smooth curve.
Step 6: Plot the points (0, 2) ;
(1, 1); (1, 3); (2, 0)
and (2, 4) on graph
sheet
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x 0 1 1 2 2
y 2 1 1 4 4
(x, y) (0, 0) (1, 1) (1, 1) (2, 4) (2, 4)
x
0 1 2 1 2y 2 3 4 1 0
(x, y) (0, 2) (1, 3) (2, 4) (1, 1) (2, 0)
Step 7: Join the points to get a line.
Step 8: From the intersecting
Curve and the line, draw
perpendiculars to the
x-axis
Step 9: Roots of the equation are x = 1 or x = 2
Example 3 : Solve the equation
Method I : x2 x 2 = 0
Split the equation
y =x2 and y = 2 +x
Step 1: Form the table of
corresponding valuesx
and y satisfying the
equation y =x2
Step 2: Form the table ofcorresponding valuesxand y satisfying theequation y = 2 +x
Step 3: Choose the scale on
x axis, 1 cm = 1 unity axis, 1 cm = 1 unit
Step 4: Plot the points (0, 0);(1, 1); (1, 1); (2, 4)and (2, 4) on the graphsheet.
Step 5: Join the points by asmooth curve
Step 6: Plot the points (0, 2);(1, 3) (2, 4); (1, 1) and
(2, 0) on the graphsheet.
Step 7: Join the points to get astraight line
Step 8: From the intersectingpoints of Curve and theline, draw the perpendi-
culars to the x-axis.
Step 9: Roots of the equation are x = 1 or x = 2
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x 0 1 1 2 2
y 2 2 0 0 4
(x, y) (0, 2) (1, 2) (1, 0) (2, 0) (2, 4)
Method II :
Step 1: Form the table of
corresponding values of
x and y satisfying
equation y =x2 x 2.
Step 2: Choose the scale on x
axis 1 cm = 1 unit and
y axis 1 cm = 1 unit.
Step 3: Plot the points (0, 2);
(1 2); (1, 0); (2, 0)
and (2, 4) on the graph
sheet.
Step 4: Join the points to form
a smooth curve
Step 5: Mark the intersecting
points of the curve and
the x axis.
Step 6: Roots of the equations are x = 1 or x = 2
Exercise : 5.9
A. 1) Draw the graph of y = x2 and find the value of 7
2) Draw the graph of y = 2x2 and find the value of 3
3) Draw the graph of y =2
1x
2 and find the value of 10
B. 1) Draw the graph of y =x2 and y = 2x+ 3 and hence solve the equation
x2 2x 3 = 0
2) Draw the graph of y = 2x2 and y = 3 x and hence solve the equation2x2 +x 3 = 0
3) Draw the graph of y = 2x2 and y = 3 +x and hence solve the equation
2x2 x 3 = 0
C. Solve graphically
1) x2 +x 12 = 0 2) x2 5x+ 6 = 0 3) x2 + 2x 8 = 0
4) x2 +x 6 = 0 5) 2x2 3x 5 = 0 6) 2x2 + 3x 5 = 0
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