Algebraic Identities and Squares - Class VIII

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

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

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

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

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

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,
$(a-b)^2=a^2-2ab+b^2$
$=(1000-2)^2$
$=(1000)^2-2(1000)(2)+(2)^2$
$=1000000-4000+4$
$=9,96,004$

$\left ( 97 \right )^{2}$
Using,
$(a-b)^2=a^2-2ab+b^2$
$=(100-3)^2$
$=(100)^2-2(100)(3)+(3)^2$
$=10000-600+9$
$=9,409$

$ \left ( 101 \right )^{2}$
Using,
$(a+b)^2=a^2+2ab+b^2$
$=(100+1)^2$
$=(100)^2+2(100)(1)+(1)^2$
$=10000+200+1$
$=10,201$

$\left ( 502 \right )^{2}$
Using,
$(a+b)^2=a^2+2ab+b^2$
$=(500+2)^2$
$=(500)^2+2(500)(2)+(2)^2$
$=2,50000+2000+4$
$=252,004$

Given, $\dfrac {x - 1}{ x} = 6$

We know the identity $a^{2}+b^{2}-2ab = (a-b)^{2}$

Given, $2l-3m= -1$

Squaring on both sides, we get 

$(2l-3m)^{2}= (-1)^{2}$

$\Rightarrow 4l^{2}+9m^{2}-12lm = 1$     .....Also given that $lm =20 $

$\Rightarrow 4l^{2}+9m^{2}-12 \times 20 = 1$

$\Rightarrow 4l^{2}+9m^{2}-240 = 1$

$\Rightarrow 4l^{2}+9m^{2}= 241$

Hence, option C is correct.

$\sqrt{12+\sqrt{12+\sqrt{12+....}}} = x$.........................(1)

$\sqrt { 3+2\sqrt { 2 }  } +\sqrt { 3-2\sqrt { 2 }  }$ 

$(i)$

Given, $a + b + c = 9$......(1) and $ab + bc + ca = 26$.......(2).

Given $f(x)=x^{2}+2(a-1)x+a+5=0$ has real roots $\implies b^{2}-4{a}{c}\geq 0 $ in $a{x^{2}}+b{x}+c=0$

$\ 83^2\(80+3)^2\= 80^2+2\cdot 80\cdot 3+3^2\= 6400+480+9\= 6889$

$x^2+\dfrac{1}{x^2}=18$

Given,

$5a\sqrt{b}-\dfrac{3}{2b\sqrt{a}}=12$

Formula,

$(a-b)^2=a^2+b^2-2ab$

squaring the given on both sides,

$25a^2b+\dfrac{9}{4b^2a}-2\left ( 5a\sqrt{b} \right )\left ( \dfrac{3}{2b\sqrt{a}} \right )=144$

$25a^2b+\dfrac{9}{4b^2a}=144+2\left ( 5a\sqrt{b} \right )\left ( \dfrac{3}{2b\sqrt{a}} \right )$

$=144+15\sqrt{\dfrac{a}{b}}$

$=144+15\sqrt{\dfrac{8b}{b}}$

$=144+15\sqrt{8}$

$\left(x+\dfrac{1}{x}\right)^3=\left(x^3+\dfrac{1}{x^3}\right)+3\left(x+\dfrac{1}{x}\right)$

$\left ( \dfrac{2x}{7} - \dfrac{7y}{4} \right ) ^{2}$

$\left ( \frac{7}{8}x + \frac{4}{5}y \right ) ^{2}$
Using,
$(a+b)^2=a^2+2ab+b^2$
$=( \frac{7}{8}x)^2+2( \frac{7}{8}x)(\frac{4}{5}y )+(\frac{4}{5}y )^2$
$=\frac{49}{64}x^2+\frac{7}{5}xy+\frac{16}{25}y^2$

Squaring,
$(3a - 4b)^2$
Using, $(a-b)^2=a^2-2ab+b^2$
$=(3a)^2-2(3a)(4b)+(4b)^2$
$=9a^2-24ab+16b^2$

Squaring,
$(\dfrac{3a}{2b} - \dfrac{2b}{3a})^2$
Using, $(a-b)^2=a^2-2ab+b^2$
$=(\dfrac{3a}{2b})^2-2(\dfrac{3a}{2b})(\dfrac{2b}{3a})+(\dfrac{2b}{3a})^2$
$=\dfrac{9a^2}{4b^2}-2+\dfrac{4b^2}{9a^2}$

Squaring,
$(a + 2b + c)^2$
Using $(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ac$
$=a^2+(2b)^2+c^2+2(a)(2b)+2(2b)(c)+2(a)(c)$
$=a^2+4b^2+c^2+4ab+4bc+2ac$

   $2a-b-3c$
Squaring, we get
   $(2a-b-3c)^2$
Using $(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ac$
$=(2a)^2+(-b)^2+(-3c)^2+2(2a)(-b)+2(-b)(-3c)+2(2a)(-3c)$
$=4a^{2} + b^{2} + 9c^{2} - 4ab + 6bc - 12ca$

    $ \left ( \dfrac{2x}{5} + \dfrac{3y}{4} - \dfrac{4z}{7} \right )^{2}$
Squaring, we get
Using $(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ac$
$=( \dfrac{2x}{5})^2+(\dfrac{3y}{4})^2+(- \dfrac{4z}{7})^2+2( \dfrac{2x}{5})(\dfrac{3y}{4})+2(\dfrac{3y}{4})(- \dfrac{4z}{7})+2( \dfrac{2x}{5})(- \dfrac{4z}{7})$
$=\dfrac{4x^{2}}{25} + \dfrac{9y^{2}}{16} + \dfrac{16z^{2}}{49} + \dfrac{3}{5} xy - \dfrac{6}{7}yz - \dfrac{16}{35}zx$

$49^2=(40+9)^2=40(40+9)+9(40+9)$
$;;;;;=(40)^2+40\times9+9\times40+9^2$
$;;;;;=1600+360+360+81$
So, $49^2=2401$
$52^2=(50+2)^2=50(50+2)+2(50+2)$
$;;;;;;=50^2+50\times2+2\times50+2^2$
$;;;;;;=2500+100+100+4$
So, $52^2=2704$.

It can be written as $43 = 40+3$
$43^2= (40+3)^2  =40^2+2\times40\times3+3^2$
$= 1600+2\times 120+  9$
$= 1849$
Therefore, D is the correct answer.

The square of $125^2$

$ = (120+5)^2$

$=14400+25+1200$

$=15625$     .....($(a+b)^2=(a^2+b^2+2ab))$

It can be written as $29 = 20+9$

On squaring both sides, we get

$29^2=(20+9)^2$

$= 20^2+2\times20\times9+9^2$

$= 400+2\times 180+  81$

$= 841$

Therefore, option A is the correct answer.

Using the formula, $a^2 - b^2 = (a + b)(a - b)$
We get, $120^2 - 119^2 = (120 + 119)(120 - 119)$
$= 239 \times 1$
$= 239$.

Using the formula, $a^2 - b^2 = (a + b)(a - b)$
We get, $36^2 - 35^2 = (36 + 35)(36 - 35)$
$= 71 \times 1$
$= 71$.

The given square $(0.98)^2$ can be evaluated as shown below:

${46}^{2} = 46\times46 = 2116$

Given term is $(3+\sqrt{2})^{5}-(3-\sqrt{2})^{5}$