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 :
$$ x = \frac {-b \pm \sqrt {b^2 -4ac}}{2a} $$
e.g. x^{2} = 9 or, x^{2} - 9 = 0
i.e. ax^{2} + c = 0 ( a ≠ 0, c = 0 )
e.g. x^{2} - 9x - 15 = 0
i.e. ax^{2} + bx + c = 0 ( a ≠ 0, b ≠ 0)
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
If the sum of two numbers is 10 and their product is 24, then what are the numbers?
The sum of digits of two digit number is 7 and their product is 12. What are the numbers?
The sum of two numbers is 16 and the sum of their squares is 130. Find the numbers.
If the sum of two numbers is 9 and their product is 18, then what are the numbers?
Find two consecutive even number whose product is 48.
The difference between the age of two sisters is 5 years and the product is 204. What is the age of the two sisters?
Sujit is 7 years older than Amisha. two years ago, the product of their ages was 18. What is their present age?
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?
The hypotenuse of a right-angled triangle is 15 cm. If the ratio of the remaining two sides is 3:4, find the two sides.
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?
ASK ANY QUESTION ON Quadratic equation
You must login to reply
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
Mar 18, 2017
0 Replies
Successfully Posted ...
Please Wait...
Gyan
If two times the number exceeds,the number formed by interchanging its digit by four ???what does it mean ??
Jan 23, 2017
0 Replies
Successfully Posted ...
Please Wait...
Hiranya
Ask any queries on this note.x^2-4x 4=0
Jan 17, 2017
0 Replies
Successfully Posted ...
Please Wait...
Ask any queries on this notex^2_4X 4=0
Jan 17, 2017
0 Replies
Successfully Posted ...
Please Wait...