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 ≠ 0, c = 0 )

ii. Adfected quadratic equation

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

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%
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If the age of father was 13 times the age of his son before 1 year.Now the age of fatherbis cube of his son age.find their present aged.

Solve please

Gyan

If two times the number exceeds,the number formed by interchanging its digit by four ???what does it mean ??

Hiranya

Ask any queries on this note.x^2-4x 4=0

Ask any queries on this notex^2_4X 4=0