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
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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
Mar 18, 2017
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Gyan
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Hiranya
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Jan 16, 2017
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Ask any queries on this notex^2_4X 4=0
Jan 16, 2017
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