Quadratic equation
An equation like ax^{2}+ bx + c = 0 where a≠ 0, which contains only one variable and '2' as its highest power is called a quadratic equation. There are two values of the variable in any quadratic equation. The roots of the equations are :
$$ x = \frac {b \pm \sqrt {b^2 4ac}}{2a} $$
Quadratic equation are of two types:
i. Pure quadratic equation
e.g. x^{2} = 9 or, x^{2}  9 = 0
i.e. ax^{2} + c = 0 ( a ≠ o, c = 0 )
ii. Adfected quadratic equation
e.g. x^{2}  9x  15 = 0
i.e. ax^{2} + bx + c = 0 ( a ≠ o, b ≠ o)
An equation like ax^{2}+ bx + c = 0 where a≠ 0, which contains only one variable and '2' as its highest power is called a quadratic equation. There are two values of the variable in any quadratic equation.. The roots of the equations are :
$$ x = \frac {b \pm \sqrt {b^2 4ac}}{2a} $$
Solution:
4x^{2} = 16x
or,4x^{2} 16x = 0
or, 4x ( x  4) = 0
Either 4x = 0
∴ x = 0
Or, x  4 = 0
∴ x = 4
Solution:
\( \frac{x^2+3}{2}\) = 6
or, \( x^ 2 + 3 \) = 12
or, \( x^2\) = 9
or, \((x)^2\) = (±3)^{2}
or, x =±3
Let, the number be x. Hence, the other number is \( x^2\)
According to question,
x + \( x^2\) = 20
or,x + \( x^2\)  20 = 0
or, \(x^2\) + 5x  4x  20 = 0
or, x ( x + 5)  4(x + 5) = 0
or, (x  4) (x + 5) = 0
Either, x  4 = 0
∴ x = 4
Or, x + 5 = 0
∴ x = 5
Let, the three numbers be x, x + 1 and x + 2
Then, According to question,
\(x ^2\) + \( (x+1)^2\) + \( (x+2)^2\) = 194
or, \(x ^2\) + \(x ^2\) + 2x + 1 + \(x ^2\) + 4x + 4 = 194
or, 3\(x ^2\) + 6x  189 = 0
or, 3(\(x ^2\) + 2x  63) = 0
or, (\(x ^2\) + 2x  63) = 0
or, \( x^2\) + 9x  7x  63 = 0
or, x (x + 9)  7( x+9) = 0
or, ( x  7) ( x + 9) = 0
Either,
x  7 = 0
∴ x = 7
Or, x + 9 = 0
∴ x = 9.
Therefore, the required numbers are 7, 7 + 1 , 7 + 2 = 7, 8, 9
Or, 9, 9 + 1, 9 + 2 = 9, 8, 7
Let, one of the odd number be x . Then, its consecutive odd number is ( x + 2)
According to question,
x ( x + 2) = 143
or, \( x^2\) + 2x = 143
or, \( x ^ 2\) + 2x  143 = 0
or, \( x ^2\) + 13x  11x  143 = 0
or, x ( x + 13)  11( x + 13) = 0
or, ( x + 13) ( x  11) = 0
Either,
x  11 = 0
or, x = 11
Or, x + 13 =0
or, x = 13
Hence, the required number is 11 and 11 + 2 = 13
Let, one of the evennumber be x . Then, its consecutive evennumber is ( x + 2)
According to question,
x ( x + 2) = 168
or, \( x^2\) + 2x = 168
or, \( x ^ 2\) + 2x  168 = 0
or, \( x ^2\) + 14x  12x  168 = 0
or, x ( x + 14)  12( x + 14) = 0
or, ( x + 14) ( x  12) = 0
Either,
x  12 = 0
or, x = 12
Or, x + 14 = 0
or, x = 14
Hence, the required number is 12 and 11 + 2 = 14
Let, one of the number be x . Then, its consecutive number is ( x + 1)
According to question,
x ( x + 1) =56
or, \( x^2\) + x =56
or, \( x ^ 2\) + x  56= 0
or, \( x ^2\) + 8x  7x  56= 0
or, x ( x + 8)  7( x + 8) = 0
or, ( x + 8) ( x  7) = 0
Either,
x  7= 0
or, x =7
Or, x + 8=0
or, x = 8
Hence, the required number is 7 and 7 + 1=8
Let, the odd number be x and its consecutive odd number be ( x + 2).
According to question,
\( (x + 2) ^2\)  \( x^2\) = 24
or,\( x^2\) + 4x + 4 \( x^2\) = 24
or, 4x = 24  4
or, 4x = 20
or, x = \( \frac {20}{4}\)
or, x = 5
Hence, the required numbers are 5 and 5 + 2 = 7
Let, the number be x.
Then, According to question,
4\( x ^ 2\) = 16x
or 4 \( x ^2\)  16x = 0
or, 4x (x  4) = 0
Either, 4x = 0
∴ x = 0
Or, x  4 = 0
∴ x = 4.,
Let, the number be x.
Then, According to question,
\( x ^ 2\)  9 = 40
or,\( x ^ 2\) = 40 + 9
or,\( x ^ 2\) = 49
or, \( (x)^2\) = \( (±7) ^ 2\)
∴x =±7
Let the whole number be x
Then, According to question,
x  10 = 39× \( \frac {1}{x}\)
or, \( x ^ 2 \)  10x = 39
or,\( x ^ 2 \)  10x  39 = 0
or,\( x ^ 2 \)  13x + 3x  39 = 0
or, x ( x  13) + 3(x  13) = 0
( x + 3) ( x  13) = 0
Either x  13 = 0
∴ x = 13
Or, x + 3 = 0
∴ x =  3
Hence, the required whole number is 13.
Let, the natural number be x
Then,
According to quesiton,
\( \frac {x ^2}{2}\)  5 = 45
or \( \frac {x ^2}{2}\) = 45 + 5
or,\( \frac {x ^2}{2}\)= 50
or, \( x ^ 2\) = 100
or, \( ( x)^2\) = \( (10)^2\) [ Only (+10) is taken because the required number is a natural number]
or, x = 10
Hence, the requierd number is 10.
Let, the first number is x then the second number becomes (12  x)
According to question,
x ( 12  x) = 32
or, \( x ^ 2\) + 12x  32 = 0
or, \( x ^ 2\)  12x + 32= 0
or, \( x ^ 2\)  8x  4x + 32 = 0
or, x ( x  8)  4( x  8 ) = 0
or, ( x  4) ( x  8) = 0
Either, x  4 = 0
∴ x = 4
Or, x  8 = 0
∴ x = 8
If first number is 4, the other number is 12  4 = 8
If first number is 8, the second number is 12  8 = 4
Hence, the required numbers are 8,4 or 4,8
Let, the age of smaller brother = x years
Then, the age of elder brother = ( x + 7) years
According to question,
x ( x + 7) = 260
or, \( x ^ 2 \) + 7x = 260
or,\( x ^ 2 \) + 7x  260 = 0
or,\( x ^ 2 \) + 20x  13x  260 = 0
or, x ( x + 20)  13 ( x + 20) = 0
or, (x+ 20) ( x  13) = 0
Either, x + 20 = 0
or, x =  20 ( Not possible)
Or,
x  13 = 0
or, x = 13
Hence, the ages of the brothers are 13 and ( 13 + 7) = 20 years
Let, the age of one sister = x years
Then, the age of other sister= yyears
From first condition, x  y = 7
or, x = y + 7 (i)
From second condition, xy = 78 (ii)
Putting the value of x in equation (ii)
(y + 7) y = 78
or, \( y ^ 2\) + 7y = 78
or, \( y ^ 2\) + 7y  78 = 0
or, \( y ^ 2\) + 13y  6y  78 = 0
or, y ( y + 13) 6( y + 13) = 0
or, ( y  6) ( y + 13) = 0
Either
y + 13 = 0
or, y = 13 (Not possible)
Or,
y  6 = 0
or, y = 6
Putting the value of y in equation (i)
x = 6 + 7 = 13
Hence, the ages of two sisters are 13 years and 6 years.
Let, x years ago, the product of their ages was 210.
Age of father x years ago = ( 35  x) years
Age of son x years ago = (12  x ) years
According to quesiton,
( 35  x) ( 12  x) = 210
or, 420  12x  35x + \( x ^ 2\) = 210
or, \( x ^ 2\)  47x + 420  210 = 0
or, \( x ^ 2\)  47x + 210 = 0
or, \( x ^2\)  42x  5x + 210 = 0
or, x ( x  42)  5( x  42) = 0
or, ( x  5) ( x  42) = 0
Either, x  5 = 0
∴ x = 5
Or, x  42 = 0
∴ x = 42 ( Not possible as their current ages are less than 42 years)
Hence, 5 years ago the product of their ages was 210.
Let, x years ago, the product of their ages was192.
Age of father x years ago = ( 42 x) years
Age of son x years ago = (16  x ) years
According to quesiton,
( 42 x) ( 16  x) =192
or, 672 42x  16x + \( x ^ 2\) =192
or, \( x ^ 2\)  58x + 672  192= 0
or, \( x ^ 2\) 58x + 480= 0
or, \( x ^2\)  48x  10x + 480 = 0
or, x ( x  48)  10( x  48) = 0
or, ( x  10) ( x  48) = 0
Either, x  10= 0
∴ x =10
Or, x  48 = 0
∴ x = 48 ( Not possible as their current ages are less than 48 years)
Hence, 10years ago the product of their ages was192.
Let, after x years the product of their ages will be 280 .
Age of elder brother after x years = ( 13 + x) years
Age of younger brother after x years = ( 7 + x) years
According to question,
( 13 + x) ( 7 + x) = 280
or, 91 + 13x + 7x + \( x ^ 2\) = 280
or, \( x ^ 2\) + 20x +91x  189 = 0
or,\( x ^ 2\) + 27x  7x  189 = 0
or, x ( x + 27)  7( x + 27 ) = 0
or, ( x + 27) ( x  7) = 0
Either, x 7= 0
∴ x =7
Or, x +27 = 0
∴ x = 27( Not possible)
Hence, after 7 years,the product of their ageswill be 280.
Let, present age of father = x years amd present age of son = y years
From first condition,
x  6 = 6( y  6)
or, x  6 = 6y  36
or, x = 6y  36 + 6
or, x = 6y  30 (i)
From second condition, xy = 300 (ii)
Putting the value of x from (i) in (ii)
( 6y  30) y = 300
or, 6\( y ^ 2\)  30y  300= 0
or, 6( \(y ^ 2\)  5y  50) = 0
or,\(y ^ 2\)  5y  50 = 0
or, \( y ^ 2\)  10y + 5y  50 = 0
or, y ( y  10) + 5 ( y  10) = 0
or, ( y + 5) ( y  10) = 0
Either, y +5 = 0
or, y = 5 ( Impossible)
Or, y  10 = 0
or, y = 10
Putting the value of y in equation (i)
x = 6 x 10  30 = 30
Hence, the present ages of father and son are 30 years and 10 years respectively.
Let, the age of elder sister = x years and the age of younger sister = y years
From first condition, xy = 150
or, y = \( \frac {150}{x}\) (i)
From second condition,
x  5 = 2 ( y  5)
or, x  5 = 2y  10
or, x = 2y  10 + 5
or, x = 2y  5 (ii)
Putting the value of y from equation (i) in equation (ii)
x = 2\( \frac {150}{x}\)  5
or, x = \( \frac { 300  5x}{x}\)
or, \( x ^ 2\) = 300  5x
or, \( x ^ 2 \) + 5x  300 = 0
or,\( x ^ 2 \) + 20x  15x  300 = 0
or, x ( x + 20)  15 ( x + 20) = 0
or, ( x  15) ( x + 20) = 0
Either x  15 = 0
∴ x = 15
Or, x + 20 = 0
∴ x =  20 ( Not possible)
Putting the value of x in euqaiton (ii)
15 = 2y  5
or, 2y = 15 + 5
or, 2y = 20
or, y = \( \frac {20}{2}\)
or, y = 10
Hence, the present ages of elder and younger sisters are 15 and 10 years respectively.
Let, the two numbers are x and y
From first condition, x + y = 21
or, y = 21  x (i)
From second condition, \( x^ 2\) + \( y ^ 2\) = 261
or, \( x ^ 2\) + \(( 21  x)^2\) = 261
or, 2\( x ^ 2\)  42x + 441 = 261
or, 2\( x ^ 2\)  42x + 180 = 0
or, 2(\( x ^ 2\)  21x + 90) = 0
or,(\( x ^ 2\)  21x + 90) = 0
or, \( x ^ 2\)  15x  6x + 90 = 0
or, x ( x  15)  6 ( x  15) = 0
or, (x  6) ( x  15) = 0
Either, x  6 = 0
∴ x = 6
Or, x  15 = 0
∴ x = 15
When, x = 15, y = 21 5 = 6
When x = 6, y = 21  6 = 15
Hence, the required numbers are 6,15 or 15,6.
Let, the two digit number is 10x + y where y is one's place and x is ten's place
From first conditon, xy = 35 (i)
From second condition, 10x + y + 18 = 10y + x
or, 10x  x + y  10y = 18
or, 9x  9y =  18
or 9 ( x  y) = 18
or, ( x  y) = \( \frac {18}{9}\)
or, x  y = 2
or, x = y  2 (ii)
Putting the value of x in equation (ii) from (i)
(y  2) y = 35
or, \( y ^ 2\)  2y = 35
or, \( y ^ 2\)  7y + 5y 35 = 0
or, y ( y  7) + 5( y  7) = 0
or, ( y + 5) ( y  7 )= 0
Either,
y  7 = 0
∴ y = 7
Or, y + 5 = 0
∴ y =  5 ( Not possible)
Putting the value of y in equation (ii)
x = 7  2 = 5
Hence, the requierd number is 10x + y = 10 x 5 + 7 = 57

If the sum of two numbers is 10 and their product is 24, then what are the numbers?
6,6
6,7
6,4
6,5

The sum of digits of two digit number is 7 and their product is 12. What are the numbers?
43,36
43,34
34,35
43,33

The sum of two numbers is 16 and the sum of their squares is 130. Find the numbers.
9 and 7
9 and 9
none
9 and 8

If the sum of two numbers is 9 and their product is 18, then what are the numbers?
6 and 5
6 and 3
6 and 4
6 and 2

Find two consecutive even number whose product is 48.
6,6
6,8
all
6,7

The difference between the age of two sisters is 5 years and the product is 204. What is the age of the two sisters?
none
12,18
13,17
12,17

Sujit is 7 years older than Amisha. two years ago, the product of their ages was 18. What is their present age?
11,3
all
11,4
11,5

A perimeter of a rectangular ground is 46 m and its area is 126 sq, meters. What is the length and breadth of the ground?
14m , 9m
15m, 9m
14m, 8m
15m, 9m

The hypotenuse of a rightangled triangle is 15 cm. If the ratio of the remaining two sides is 3:4, find the two sides.
11 cm, 9cm
13cm, 9cm
9 cm, 12 cm
9 cm , 11 cm

Rs.120 is equally divided among a certain number of students. If there were 3 students more, each would have received Rs. 2 less. What is the numbers of students?
13
15
12
14

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