Tag: finding the square of a number

$(5+\sqrt3)^2\=5^2+(\sqrt3)^2+2\times5\times\sqrt3\=25+3+10\sqrt3\=28+10\sqrt3$

$(\sqrt5+\sqrt6)^2\=(\sqrt5)^2+(\sqrt6)^2+2\times \sqrt5\times\sqrt6\=5+6+2\sqrt{30}\=11+2\sqrt{30}$

$(5-6\sqrt3)^2\=5^2+(6\sqrt3)2-2\times5 \times6 \sqrt3\=25+108-60\sqrt3\=133-60\sqrt3$

$(3+2\sqrt5)^2\=3^2+(2\sqrt5)^2+2\times3\times2\sqrt5\=9+4\times5+12\sqrt5\=29+12\sqrt5$

$[(5^2+12^2)^{(\frac{1}{2})}]\times3\=(25+144)^{(\frac{1}{2})}\times3\=169^{(\frac{1}{2})}\times3\=(13^2)^{(\frac{1}{2})}\times3\=13\times3=39$

Squaring,
$(2a+b)^2$
$=(2a)^2+2(2a)(b)+(b^2)$
$=4a^2+4ab+b^2$

Squaring,
$(3a + 7b)^2$
$=(3a)^2+2(3a)(7b)+(7b)^2$
$=9a^2+42ab+14b^2$

To find, value of : $\sqrt{3\,+\,2\,\sqrt{2}}\,-\,\sqrt{3\,-\,2\,\sqrt{2}}$ 
Let $x = \sqrt{3\,+\,2\,\sqrt{2}}\,-\,\sqrt{3\,-\,2\,\sqrt{2}}$ 
Squaring both sides:
$x^2$ = $\displaystyle\,\left ( \sqrt{3\,+\,2\sqrt{2}}\,-\,\sqrt{3\,-\,2\sqrt{2}} \right )^{2}$
$\displaystyle\,x^2\,=\,3\,+\,2\sqrt{2}\,+\,3\,-\,2\sqrt{2}\,-\,2(3\,+\,2\sqrt{2})(3\,-\,2\sqrt{2})$
$\Rightarrow x^2\,=\,4$
$\Rightarrow x\,=\,2$
Hence, option 'A' is correct.

$\left ( 998 \right )^{2}$
Using,
$(a-b)^2=a^2-2ab+b^2$
$=(1000-2)^2$
$=(1000)^2-2(1000)(2)+(2)^2$
$=1000000-4000+4$
$=9,96,004$