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Title: Algebra _ Problems leading to quadratic equations
Description: Algebra _ Problems leading to quadratic equations Many word problems Involving unknown quantities can be translated for solving quadratic equations Methods of solving quadratic equations are discussed h ere in the following steps. Step I: Denote the unknown quantities by x, y etc. Step II: use the conditions of the problem to establish in unknown quantities. Step III: Use the equations to establish one quadratic equation in one unknown. Step IV: Solve this equation to obtain the value of the unknown in the set to which it belongs.
Description: Algebra _ Problems leading to quadratic equations Many word problems Involving unknown quantities can be translated for solving quadratic equations Methods of solving quadratic equations are discussed h ere in the following steps. Step I: Denote the unknown quantities by x, y etc. Step II: use the conditions of the problem to establish in unknown quantities. Step III: Use the equations to establish one quadratic equation in one unknown. Step IV: Solve this equation to obtain the value of the unknown in the set to which it belongs.
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Algebra
Problems leading to quadratic equations
1
...
Solution:
Let the two consecutive positive odd numbers be a and a + 2
...
The sum of a number and its reciprocal is 2
Solution:
1
Find the number
...
Given
x+
1
x
x2 + 1
x
30x2 + 30
= 2
1
30
=
61
30
= 61x
30x2 – 61x + 30
= 0
30x2 – 36x – 25x + 30
= 0
6x (5x – 6) –5(5x – 6) = 0
(5x – 6)(6x – 5) = 0
x = 6 or 5
5
6
3
...
Solution:
n (n + 1)
2
S =
231 =
Given S = 231
n (n + 1)
2
n ( n + 1) = 462
n2 + n – 462 = 0
n2 + 22n 21n 462 = 0
n(n + 22) 21(n + 22) = 0
(n + 22)( n 21) = 0
n = 22
or n = 21
n = 21 (as n = –22 in not a negative number)
4
...
Find the multiples
...
5
...
Solution:
Given: a + b = 16 (a and b are two part)
b = 16 a
…(1)
2a2 = b2 + 164
2a2 = (16 a)2 + 164
= 256 32a + a2 + 164
a2 + 32a 420 = 0
a2+ 42a 10a 420 = 0
a (a + 42 ) –10 (a + 42) = 0
(a + 42) (a 10) = 0
a
= 42 (not possible)
a
= 10
(1) b = 6
The parts are 10 and 6
6
...
The product of the digits is 24 and the difference between the digits is
5
...
Solution:
Let the number be xy
...
A two digit number is such that the product of the digits is 18
...
Find the number
...
8
...
Solution:
By pythagorous theorem
(2x+1)2
4x2 + 4x + 1
4x2 8x
= (2x1)2 + (2x)2
= 4x2 4x + 1 + 4x2
= 0
x2 2x = 0
x(x 2) = 0
x
= 0 or x = 2 (x = 0 is not possible)
x = 2 is taken
9
...
Five years ago, the
product of their ages (in years ) was 124
...
Let x be the father’s age
Solution:
Let y be the son’s age
Given:
x + y = 45
…(1)
(x 5)( y 5) = 124
xy 5x 5y + 25 = 124
From (1)
x
= 45 y
(45y) y 5(45 y) 5y + 25 = 124
45y – y2 – 225 + 5y 5y + 25 = 124
y2 + 45y 324
= 0
y2 45y + 324
= 0
(y 9)(y 36)
= 0
y
y
= 9 or 36 ( y = 36 is not possible)
= 9 x = 36
The present age of father is 36
The present age of son is 9
10
...
cm
...
Solution:
Let l be the length, b be the breadth of the rectangle
2l + 2b
lb
= 36 l + b = 18 b =18 l
= 80
l (18 – l ) = 80
18l l2
= 80
l2 18l + 80 = 0
(l 10)(l 8) = 0
l
= 10 or l = 8
b
= 8 or b = 10
Length = 10cm, breadth = 8cm
11
...
Determine its usual speed
...
1500
1
+
x + 250
2
1500
x
=
1500
x
=
(1500)
...
12
...
The square of
the smaller number is four times the larger number
...
Solution :
Let the number be a
Given:
a2 b2
Also
b2
a2 4a
2
a 4a 45
(a 9)(a + 5)
a
and b
= 45
= 4a
= 45
= 0
= 0
= 9 or
a
= 9
b
= 6
a = 5 (not possible)
and b2 = 4 x 9 = 36
The numbers are 9 and 6
Title: Algebra _ Problems leading to quadratic equations
Description: Algebra _ Problems leading to quadratic equations Many word problems Involving unknown quantities can be translated for solving quadratic equations Methods of solving quadratic equations are discussed h ere in the following steps. Step I: Denote the unknown quantities by x, y etc. Step II: use the conditions of the problem to establish in unknown quantities. Step III: Use the equations to establish one quadratic equation in one unknown. Step IV: Solve this equation to obtain the value of the unknown in the set to which it belongs.
Description: Algebra _ Problems leading to quadratic equations Many word problems Involving unknown quantities can be translated for solving quadratic equations Methods of solving quadratic equations are discussed h ere in the following steps. Step I: Denote the unknown quantities by x, y etc. Step II: use the conditions of the problem to establish in unknown quantities. Step III: Use the equations to establish one quadratic equation in one unknown. Step IV: Solve this equation to obtain the value of the unknown in the set to which it belongs.