Quadratic equation

An equation like ax2+ 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. x2 = 9 or, x2 - 9 = 0
i.e. ax2 + c = 0 ( a ≠ o, c = 0 )

ii. Adfected quadratic equation

e.g. x2 - 9x - 15 = 0
i.e. ax2 + bx + c = 0 ( a ≠ o, b ≠ o)

An equation like ax2+ 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:

4x2 = 16x

or,4x2- 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

0%
  • If the sum of two numbers is 10 and their product is 24, then what are the numbers?

    6,6
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    43,36
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    9 and 7
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    6 and 5
    6 and 3
    6 and 4
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    6,6
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    none
    12,18
    13,17
    12,17
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    11,3
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    11,5
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    14m , 9m
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    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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