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SECOND DEGREE(E) Flipbook PDF
SECOND DEGREE(E)
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SECOND DEGREE EQUATIONS * General form = ax2 + bx +c = 0, a≠ 0
* If x2 = k ,then x = ± √k
* If (x+a)2 = k ,then x+a = ± √k
* If (x+a)(x-b) = 0 ,then x+a = 0 or (x-b)=0
a * To become x2 + ax is a perfect square, we have to add ( 2 ) 2 to it * 5 is added to the number = x + 5
* 6 is subtracted from the number = x – 6
* Two times of a number = 2x
* Half of a number =
* One third of a number = x 3
* Square of a number = x2
x 2
* 5 is added to 3 times of a number = 3x + 5 * 2 is subtracted from 4 times of a number = 4x – 2 * 3 is added to square of a number = x2 + 3 * 2 is subtracted from the square of a number = x2 - 2 * 4 times of a number is added to the square of a number = x2 + 4x * 2 times of a number is subtracted from the square of a number = x2 - 2x * 2 is added to the number,the result will be squared = (x+2)2 * 3 is subtracted from the number,the result will be squared = (x-3)2 * Four times the square of a number = 4x2 * Consecutive 2 natural numbers (Integers,counting numbers) = x,x+1 * Alternate 2 natural numbers (Integers,counting numbers) = x,x+2 * Consecutive 2 odd numbers (2 even numbers) = x,x+2 * Alternate 2 odd numbers ( 2 even numbers ) = x,x+4 * Consecutive multiples of 3 = x, x+3 * The consecutive numbers with common difference 5 = x,x+5 * If sum of two numbers 8, then numbers = 4 + x,4 – x or x,8 – x * If difference of two numbers = 10 , then numbers=x,x+10
or x , x - 10
JITHESH P / HST MATHS / GGVHSS WANDOOR
* Rectangle Perimeter = 2 ( length + breadth) , Area = length × breadth Length2 + breadth2 = Hypotenuse 2 * Square Area = a2
Perimeter = 4a ,
Length of the diagonal = a√2
* Triangle Area = 1 bh ( b = length of one side , h= height of the distance from opposite 2 vertex to that side.) * Right angled triangle 1 Area = bh [ b = base and h = Altitude ] 2
, Base2 + Altitude2 = Hypotenuse 2
* Regular polygon Sum of interior angle = (n-2) × 180 Number of diagonals =
Sum of exterior angle = 3600
n(n-3) 2
* Compound interest A = P( 1 + R ) N 100
, Where A = Final amount , P = initial principal value
R = Interet rate , N = Number of years PROBLEMAS ( 1) Solve the following : (a) x2 = 36
(2) (x – 3)2 = 100
(d) x2 + 8x + 16 = 144
(e) x2 - 10x + 25 = 64
(c) (2x+3)2 = 81
Answer : (a) x2 = 36 then x = ± √36 = ± 6 (b) (x – 3)2 = 100 (x – 3) = ± √100 then x – 3 = ± 10 x – 3 = 10
or x – 3 = - 10
x = 10 + 3 = 13
or x = -10 + 3 = - 7
JITHESH P / HST MATHS / GGVHSS WANDOOR
(c) (2x+3)2 = 81 2x + 3 = ± √81 then 2x + 3 = ± 9 2x + 3 = 9
or 2x + 3 = -9
2x = 9 – 3
or 2x = -9 - 3
2x = 6
or 2x = -12
x=6=3 2 2 (d) x + 8x + 16 = 144
or x = -12 = -6 2
(x + 4)2 = 144 x + 4 = ± √144 x + 4 = ± 12 x + 4 = 12
or x + 4 = -12
x = 12 – 4 = 8
or x = -12 – 4 = -16
(e) x2 - 10x + 25 = 64 (x - 5)2 = 64 x - 5 = ± √64 x-5=±8 x-5=8
or x – 5 = -8
x = 8 + 5 = 13
or x = -8 + 5 = -3
(2)
Take side of the first square = x Side of the second square = x + 1 (x+ 1)2 = 36 x + 1 = √36 x+1=6
, x = 6 – 1 = 5 , Side of the original square = 5 meter
JITHESH P / HST MATHS / GGVHSS WANDOOR
(3)
(4)
WORK SHEET ( 1) Solve the following : (a) x2 = 81
(2) (x – 7)2 = 121
(d) x2 + 6x + 9 = 256
(e) x2 - 12x + 36 = 32
(c) (3x-2)2 = 49
(2)
JITHESH P / HST MATHS / GGVHSS WANDOOR
SECOND DEGREE EQUATIONS 1
The square of a term in the arithmetic sequence 2,5,8.........,is 2500.What is its position ?
Sequence = 2,5,8,..... First term (f) = 2
Common difference (d) = 5 – 2 = 3
nth term = dn + f – d = 3n + 2 – 3 = 3n – 1 Square of a term = 2500 (3n – 1)2 = 2500 3n – 1 = √2500 3n – 1 = 50 3n = 50 + 1 3n = 51 n = 51 = 17 3 Position of the term = 17
2
P = 2000
N = 2 and A = 2205
R 2 ) =A 100 R 2 2000 (1 + ) = 2205 100 R (1 + )2 = 2205 100 2000 R 2 441 (1 + ) = 100 400 R 441 1+ = 100 400 R = 21 - 1 100 20 1 R = 100 × 1 = 20 P( 1 +
21 20 21 - 20 = = 1 20 20 100 = 5 , Rate of interest = 5 % 20 =
JITHESH P / HST MATHS / GGVHSS WANDOOR
3
Find the length of the side of the green square Take side of the large square = x meter Area of the square = x2 Area of 2 yellow rectangle = 2 × x × 1 = 2x
x
Area of the small square = 12 = 1 x
Total area = 100
1 1
x2 + 2x + 1 = 100 (x+1)2 = 100
x
x + 1 = √ 100 = 10 x = 10 – 1 = 9 Length of the side of the green square = 9 meter
x
1
WORKSHEET
JITHESH P / HST MATHS / GGVHSS WANDOOR
1
1
SECOND DEGREE EQUATIONS 1
Take, length = x , then breadth = x-2 Area = 224 x(x-2) = 224 x2 – 2x = 224 x2 – 2x + 1 = 224 + 1 ( x – 1)2 = 225
Completing the square method Coefficient of x = -2 Half =
-2 = -1 2
Square of the half = (-1)2 = 1 Adding 1 to both side of the equation
x – 1 = √ 225 x – 1 = 15 x = 15 + 1 = 16 length = x = 16 meter and Breadth = x – 2 = 16 – 2 = 14
2
Consecutive even numbers = x , x+2 x(x+2) + 1 = 289 x2 + 2x + 1 = 289 (x + 1)2 = 289 x + 1 = √ 289
Completing the square method Coefficient of x = 2 Half =
2 = 1 2
Square of the half = (1)2 = 1 Adding 1 to both side of the equation
x + 1 = 17 x = 17 – 1= 16 Consecutive numbers = x , x + 2 = 16 , 16 + 2 = 16 , 18
JITHESH P / HST MATHS / GGVHSS WANDOOR
3. 3
A rectangle is to be made with perimeter 100 meters and area 525 square meters. What should be the length of its sides. Perimeter = 100 2 ( length + breadth ) = 100 length + breadth = 100 = 50 2 Take length = 25 + x and breadth = 25 – x Area = 525 ( 25 + x )( 25 – x ) = 525 252 – x2 = 525 625 – x2 = 525 - x2 = 525 – 625 - x2 = -100 x2 = 100 x = √ 100 = 10 length = 25 + x = 25 + 10 = 35 meter and breadth = 25 – x = 25 – 10 = 15 meter Worksheet
1.
The difference of two positive numbers is 6 and their product is 216.What are the numbers.
2.
9 added to the product of two consecutive multiples od 6 gives 729.What are the numbers.
3.
Using completing the square method,find the values of x ? (a) x2 + 8x = 9
(b) x2 - 12x = -35
(c) x2 + 5x = 24
(d) x2 + 3x = 40
(e) x2 + 10x + 16 = 0
(f) x2 - 5x - 36 = 0
JITHESH P / HST MATHS / GGVHSS WANDOOR
SECOND DEGREE EQUATIONS 1.
How many terms of the arithmetic sequence 5,7,9,.....must be added to get 140 ? First term ( f ) = 5
Common difference ( d) = 7 – 5 = 2
{p= d = 2=1 , 2 2
q=f- d =5–1=4 2 2 Sum of the first n terms = pn + qn = n2 + 4n Given that , Sum = 140 n2 + 4n = 140 n2 + 4n + 4 = 140 + 4 (n + 2)2 = 144 n + 2 = √144 = 12
An isosceles triangle has to made like this. Height should be 2 meters less than the base and the area of the triangle should be 12 square meters.What
Base
should be the length of its sides.? Take base = x , then height = x – 2 1 x(x-2) = 12 2 x(x-2) = 12 × 2 = 24
A
x2 – 2x = 24
6–2=4m
(2)
Height
Number of terms = 12
x2 – 2x + 1 = 24 + 1 (x -1)2 = 25 x – 1 = √25 = 5
B
C 3m
x=5+1=6 AB = AC = √ 32 + 42 = √ 9 + 16 = √ 25 = 5 cm and BC = 6 cm JITHESH P / HST MATHS / GGVHSS WANDOOR
3m
WORK HEET 1.
A 2.6 meter lon rod leans against a wall ,its foot 1 meter from the wall.When foot is moved a little away from the wall,its upper end slides the same length down.How
d Ro
Wall
much farther is the water moved ?
Ground 2.
16 added to the sum of the first few terms of the arithmetic sequence 9, 11, 13, ... gave 256. How many terms were added?
JITHESH P / HST MATHS / GGVHSS WANDOOR
SECOND DEGREE EQUATIONS 1
The product of a number and 4 more than that is 480, what are the numbers. Take , numbers = x , x+4
x (x+4) = 480
x2 + 4x = 480 x2 + 4x + 4 = 480 + 4 (x + 2)2 = 484 x + 2 = ±√484 x + 2 = ±22 x + 2 = 22
or x + 2 = - 22
x = 22 – 2 = 20
or x = -22 – 2 = - 24
Number = x,x+4 = 20,20 + 4 = 20 , 24 Number = x , x+4 = -24 , -24 + 4 = -24, -20 2.
Find two numbers with sum 6 and product -27. Sum = 6 , numbers = 3 + x , 3 – x (3 + x) (3-x) = -27 32 – x2 = -27 9 – x2 = -27
JITHESH P / HST MATHS / GGVHSS WANDOOR
- x2 = - 27 – 9 -x2 = -36 x2 = 36
x = ±√36 x = ±6
,
Numbers = 3 + x , 3 – x = 3 + 6, 3 – 6 = 9 , -3 Number = 3 + x , 3 – x = 3 + -6 , 3 - -6 = -3 , 3 + 6 = -3,9 WORK SHEET 1.
The product of a number and 2 more than that is 168, what are the numbers.
2.
Find two numbers with sum 4 and product -77.
JITHESH P / HST MATHS / GGVHSS WANDOOR
SECOND DEGREE EQUATIONS (1)
Find two numbers with sum 4 and product 2. Sum = 4 , numbers = 2 + x , 2 – x (2 + x) (2-x) = 2 22 – x2 = 2 4 – x2 = 2 - x2 = 2 - 4 -x2 = -2 x2 = 2 x = ±√2 , Numbers = 2+ x , 2 – x = 2+ √2 , 2 - √2 Number = 2 + x , 2 – x = 2 - √2 , 2 + √2
(2)
How many terms of the arithmetic sequence 99, 97, 95, must be added to get 900? Sequence = 99,97,95,......... First term(f) = 99 , Common difference (d) = 97 – 99 = -2 p = d = -2 = -1 2
2
q = f - d = 99 - -1 = 99 + 1 = 100 2
Sum of the first ‘ n’ terms = pn2 + qn = - n2 + 100 n - n2 + 100 n = 900 n2 – 100n = -900
{ multiplying through out by - 1 }
n2 – 100n = -900 n2 – 100n + 2500 = -900 + 2500
JITHESH P / HST MATHS / GGVHSS WANDOOR
(n – 50 ) 2 = 1600 n – 50 = ±√1600 = ±40 n – 50 = 40
or n – 50 = - 40
n = 40 + 50
or n = -40 + 50
n = 90
or n = 10
Number of terms = 90 , 10 A rod 28 centimetres long is to be bent to make a rectangle. (i) Can a rectangle of diagonal 8 centimetres be made? (ii) Can a rectangle of diagonal 10 centimetres be made? (iii) How about a rectangle of diagonal 14 centimetres? Calculate the lengths of the sides of the rectangles that can be made. (i) Perimeter = 28 2 ( length + breadth) = 28
8
length + breadth = 28 = 14 2
sides = 7 + x , 7 – x
7-x
(3)
7+x
length2 + breadth2 = Diagonals2 (7 + x)2 + (7 – x)2 = 82 72 + 2 × 7 × x + x2 + 72 - 2 × 7 × x + x2 = 64 49 + 14x + x2 + 49 – 14 x + x2 = 64 2x2 + 98 = 64 2x2 = 64 – 98 2x2 = -34 x2 = -34 = - 17 2
The square of any term cannot be negative. So we can’t form a rectangle.
JITHESH P / HST MATHS / GGVHSS WANDOOR
(ii) Perimeter = 28 10
7-x
2 ( length + breadth) = 28 length + breadth = 28 = 14 2
sides= 7 + x , 7 – x
7+x
length2 + breadth2 = Diagonals2 (7 + x)2 + (7 – x)2 = 102 72 + 2 × 7 × x + x2 + 72 - 2 × 7 × x + x2 = 100 49 + 14x + x2 + 49 – 14 x + x2 = 100 2x2 + 98 = 100 2x2 = 100 – 98 2x2 = 2 x2 = 2 = 1 2
x = ±√1 =± 1 Length of the sides = 7 + x , 7 – x = 7 + 1 , 7 – 1 = 8 cm , 6 cm Length of the sides = 7 + x , 7 – x = 7 + -1 , 7 – -1 = 6 cm , 8 cm
2 ( length + breadth) = 28
14
length + breadth = 28 = 14 2
sides= 7 + x , 7 – x
7+x
length2 + breadth2 = Diagonals2 (7 + x)2 + (7 – x)2 = 142 72 + 2 × 7 × x + x2 + 72 - 2 × 7 × x + x2 = 196 49 + 14x + x2 + 49 – 14 x + x2 = 196 2x2 + 98 = 196
JITHESH P / HST MATHS / GGVHSS WANDOOR
7-x
(iii) Perimeter = 28
2x2 = 196 – 98 2x2 = 98 x2 = 98 = 49 2
x = ±√49 =± 7 Length of the sides = 7 + x , 7 – x = 7 + 7 , 7 – 7 = 14 cm , 0 cm Length of the sides = 7 + x , 7 – x = 7 + -7 , 7 – -7 = 0 cm , 14 cm. So,we can’t form a rectangle with one of the side 0 cm. WORKSHEET (1)
A rectangle of width 8 centimeters is cut off from a square sheet along its side. The remaining portion has an area 84 cm2 . Calulate the side of the square ?
(2)
The length of the rectangle is 3 meter more than 3 times its breadth.Its diagonal is 1 meter more than the length. Find the length and breadth of the rectangle ?
JITHESH P / HST MATHS / GGVHSS WANDOOR
SECOND DEGREE EQUATIONS The general form of a second degree equation is ax2 + bx + c = 0 where a≠0 Solutions are ,
x=
-b ± √b2 – 4ac 2a
(1) Solve 3x2 + 2x - 1 = 0
WORKSHEET (1)
Solve the following equations : (a) x2 – 9x + 20 = 0
(b) x2 – 11x - 26 = 0
(c) 6x2 + 5x - 6 = 0
(d) 3x2 + 14x - 5 = 0
(e) 2x2 – 7x + 6 = 0
(f) 2x2 – 9x - 5 = 0
JITHESH P / HST MATHS / GGVHSS WANDOOR
SECOND DEGREE EQUATIONS 1
1 + 2 + 3 + .................+ n = 300 n(n+1) = 300 2
n(n+1) = 2 × 300 n2 + n = 600 n2 + n – 600 = 0 a = 1 , b = 1 , c = -600 x=
-b ± √b2 – 4ac 2
-1 ± √12 – 4 ×1 × -600 x= 2 -1 ± √12 + 2400 x= 2 x = -1 ± √2401 = -1 ± 49 2 2 -1 + 49 48 x= = 2 = 24 2
Number of terms = 24 2
Wrong perimeter = 24 2( length + breadth ) = 24 2( 10 + breadth) = 24 10 + breadth = 24 2
10 + breadth = 12 breadth = 12 – 10 = 2 Area = length × breadth = 10 × 2 = 20 m2
JITHESH P / HST MATHS / GGVHSS WANDOOR
Correct perimeter = 42 and
correct Area = 20
2( length + breadth ) = 42 length + breadth = 42 = 21 2
Sides = 10.5 + x , 10.5 – x (10.5 + x)(10.5 – x) = 20 10.52 – x2 = 20 110.25 – x2 = 20 – x2 = 20 – 110. 25 – x2 = - 90.25 x2 = 90.25 x = √90.25 = 9.5 Length of the sides = 10.5 + x , 10.5 – x = 10.5 + 9.5 ,10.5 – 9.5 = 20 meter ,1 meter 3
The wrong answers are
x=4,x=6
ie, x – 4 = 0 and x – 6 = 0 (x – 4)(x-6) = 0 x2 – 6x – 4x + 24 = 0 x2 – 10x + 24 = 0 The wrong second degree equation is , x2 – 10x + 24 = 0 The Correct second degree equation is , x2 – 10x – 24 = 0
JITHESH P / HST MATHS / GGVHSS WANDOOR
x2 – 10x – 24 = 0 x2 – 10x = 24 x2 – 10x + 25 = 24 + 25 ( x – 5)2 = 49 x -5 = ±√ 49 x–5 =±7 x–5=7
Or
x – 5 = -7
x = 7 + 5 = 12
Or
x = - 7 + 5 = -2
The correct answers are 12 , -2 WORKSHEET 1 2
3
To get more question , click the following link Second degree equations (Question)
JITHESH P / HST MATHS / GGVHSS WANDOOR