Tag: maths

Repeated subtraction method: Subtract successive odd numbers from the given number starting from 1 till the difference becomes zero.
So, 64 - 1 = 63
63 - 3 = 60
60 - 5 = 55
55 - 7 = 48
48 - 9 = 39
39 - 11 = 28
28 - 13 = 15
15 - 15 = 0.
15 is the odd number, we get the value of zero for the square root of the number 64.

$\sqrt{\dfrac{144}{256}}$
= $\dfrac{12}{16}$.
= 0.75

$\sqrt{1369}= \sqrt{37 \times 37}$
So, the square root of 1369 is 37.

Repeated subtraction method: Subtract successive odd numbers from the given number starting from 1 till the difference becomes zero.
So, 169 - 1 = 168
168 - 3 = 165
165 - 5 = 160
160 - 7 = 153
153 - 9 = 144
9 is the odd number, subtracting with 153 to get the value of 144.

$\sqrt{\dfrac{784}{196}}$
= $\dfrac{28}{14} = 2$.

Let $A=\sqrt { x } +\cfrac { 1 }{ \sqrt { x }  } $
$\Rightarrow { A }^{ 2 }=x+\cfrac { 1 }{ x } -2=\left( 5+2\sqrt { 6 }  \right) +\cfrac { 1 }{ 5+2\sqrt { 6 }  } -2$
$=5+2\sqrt { 6 } +\cfrac { 5-2\sqrt { 6 }  }{ 25-24 } -2=8$ $\text{[Rationalising the denominator]}$
$=5+2\sqrt { 6 } + { 5-2\sqrt { 6 }  } -2=8$
$ \Rightarrow A^2=8 $
$ \Rightarrow A=2\sqrt { 2 } $

${ ax }^{ 2 }+2b{ |x| }-c=a{ |x| }^{ 2 }+2b|x|-c$

Rupee collected from each member $=$ number of member in group $=$ $k$

If $\displaystyle x+\frac { 1 }{ 4 } \sqrt { x } +{ a }^{ 2 }$ is a perfect square
then $\displaystyle \frac { 1 }{ 4 } \sqrt { x } =2\times \sqrt { x } \times \left( \pm a \right) $
$\displaystyle \therefore \quad a=\pm \frac { 1 }{ 8 } $