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Finding the square of a number - class-VIII

Description: finding the square of a number
Number of Questions: 55
Created by:
Tags: square and square roots maths squares and square roots squares, square roots, cubes, cube roots square roots and cube roots
Attempted 0/54 Correct 0 Score 0

Write down the values of:
$(5+\sqrt3)^2$

  1. $22-10\sqrt{3}$

  2. $28+10\sqrt{3}$

  3. $22+10\sqrt{3}$

  4. $28-10\sqrt{3}$


Correct Option: B
Explanation:

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

Write down the values of:
$(\sqrt5+\sqrt6)^2$

  1. $\sqrt5+\sqrt6+\sqrt{30}$

  2. $11+2\sqrt{30}$

  3. $11+\sqrt{30}$

  4. $\sqrt{11}+60$


Correct Option: B
Explanation:

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

Expand $(5-6\sqrt3)^2$

  1. $133+30\sqrt3$

  2. $133-60\sqrt3$

  3. $83-60\sqrt3$

  4. $83+30\sqrt3$


Correct Option: B
Explanation:

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

Write down the values of:
$4(\sqrt6-3)^2$

  1. $30-24\sqrt{6}$

  2. $15-6\sqrt{6}$

  3. $15+12\sqrt{6}$

  4. $60-24\sqrt{6}$


Correct Option: D
Explanation:
$4(\sqrt6-3)^2\\=4[(\sqrt6)^2+3^2-2\cdot \sqrt 6\cdot 3]\\4[6+9-6\sqrt6]\\=4(15-6\sqrt6)\\=12(5-2\sqrt6)\\=60-24\sqrt6$

Write down the values of:
$(3+2\sqrt5)^2$

  1. $29+12\sqrt{5}$

  2. $29+6\sqrt{5}$

  3. $19-12\sqrt{5}$

  4. $29-6\sqrt{5}$


Correct Option: A
Explanation:

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

Evaluate: ${(5^2+12^2)^{\frac{1}{2}}}3$

  1. $54$

  2. $34$

  3. $39$

  4. $59$


Correct Option: C
Explanation:

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

Find the square of  $2a+b$

  1. $4a^{2} + 4ab + b^{2}$

  2. $a^{2} + ab + b^{3}$

  3. $4a^{2} + ab + b^{2}$

  4. $4a^{2} + 4ab - b^{2}$


Correct Option: A
Explanation:

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

Find the square of $3a + 7b$

  1. $9a^{2} + 42ab + 49b^{2}$

  2. $9a^{2} + 40ab + 49b^{2}$

  3. $18a^{2} + 42ab + 98b^{2}$

  4. $18a^{2} + 40ab + 98b^{2}$


Correct Option: A
Explanation:

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

$\sqrt{3\,+\,2\,\sqrt{2}}\,-\,\sqrt{3\,-\,2\,\sqrt{2}}$ is equal to 

  1. $2$

  2. $1$

  3. $2\sqrt{2}$

  4. $\sqrt{6}$


Correct Option: A
Explanation:

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.

Use identities to evaluate $\displaystyle \left ( 998 \right )^{2}$

  1. $\displaystyle 9,96,004$

  2. $\displaystyle 9,16,004$

  3. $\displaystyle 9,96,104$

  4. $\displaystyle 9,96,324$


Correct Option: A
Explanation:

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

Use identities to evaluate $\displaystyle \left ( 97 \right )^{2}$

  1. $\displaystyle 9,109$

  2. $\displaystyle 9,409$

  3. $\displaystyle 9,909$

  4. $\displaystyle 9,209$


Correct Option: B
Explanation:

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

Use identities to evaluate $\displaystyle \left ( 101 \right )^{2}$

  1. $\displaystyle 10,301$

  2. $\displaystyle 12,301$

  3. $\displaystyle 10,201$

  4. $\displaystyle 12,201$


Correct Option: C
Explanation:

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

Use identities to evaluate :$\displaystyle \left ( 502 \right )^{2}$

  1. $\displaystyle 1,62,004$

  2. $\displaystyle 1,22,004$

  3. $\displaystyle 2,12,004$

  4. $\displaystyle 2,52,004$


Correct Option: D
Explanation:

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

The simplified value of $\displaystyle \left ( \sqrt{3}+1 \right )^{2}-2\left ( 2+\sqrt{3} \right )$ is

  1. 2

  2. -1

  3. 1

  4. 0


Correct Option: D
Explanation:
$(\sqrt{3}+1)^{2}-2(2+\sqrt{3})$
=$(3+1+2\sqrt{3})-2(2+\sqrt{3})$
=$4+2\sqrt{3}-4-2\sqrt{3}=0$
Evaluate the following 
$(97)^{2}$
  1. $9409$

  2. $9049$

  3. $9949$

  4. $4949$


Correct Option: A
Explanation:
$ (97)^{2} = (100-3)^{2} = 100^{2}+3^{2}-2(400)(3) $
$ = 10000+9-600 = \boxed {9409} $ 

If $3x - \dfrac {1}{2x} = 6$, then the value of $9x^{2} + \dfrac {1}{4x^{2}}$

  1. $36$

  2. $33$

  3. $30$

  4. $39$


Correct Option: D
Explanation:
Given that $3x-\dfrac{1}{2x} = 6$
Squaring both sides, we get
$9x^{2}+\dfrac{1}{4x^{2}}-3 = 36$
We know the identity $a^{2}+b^{2}-2ab = (a-b)^{2}$
$9x^{2}+\dfrac{1}{4x^{2}}= 39$     

If $x - \dfrac {1}{x} = \sqrt {6}$, then $x^{2} + \dfrac {1}{x^{2}}$ is ________.

  1. $2$

  2. $4$

  3. $6$

  4. $8$


Correct Option: D
Explanation:

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

Multiplying and divide the above equation with $x - \dfrac {1}{x}$
Thus $ \dfrac{x-\dfrac{1}{x}\times x-\dfrac{1}{x}}{x-\dfrac{1}{x}} = \sqrt{6} $
Using $(a-b)^{2} = a^{2} + b^{2} - 2ab $
and substituting $x-\dfrac{1}{x} = \sqrt{6}$  in denominator, we get
$\dfrac{x^{2} + \dfrac{1}{x^{2}} - 2x\dfrac{1}{x}}{\sqrt{6}} = \sqrt{6}$
$\Rightarrow x^{2} + \dfrac{1}{x^{2}} - 2  =\sqrt{6}\times \sqrt{6}$
$\Rightarrow x^{2} + \dfrac{1}{x^{2}} = 6+2 = 8$

If $2l - 3m = -1$ and $lm = 20$, then the value of $4l^{2} + 9m^{2}$ is ________.

  1. $239$

  2. $240$

  3. $241$

  4. $361$


Correct Option: C
Explanation:

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.

On simplification the product of given expression $\left (x - \dfrac {1}{x}\right )\left (x + \dfrac {1}{x}\right )\left (x^{2} + \dfrac {1}{x^{2}}\right )$ is ________.

  1. $x^{3} - \dfrac {1}{x^{3}}$

  2. $x^{3} + \dfrac {1}{x^{3}}$

  3. $x^{4} - \dfrac {1}{x^{4}}$

  4. $x^{4} + \dfrac {1}{x^{4}}$


Correct Option: C
Explanation:
We need to find value of $\left (x-\dfrac{1}{x}\right)\left (x+\dfrac{1}{x}\right)\left (x^{2}+\dfrac{1}{x^{2}}\right)$
We know the identity $a^{2}-b^{2} = (a-b)(a+b)$

Then applying this to first two terms:

$\left (x^{2}-\dfrac{1}{x^{2}}\right)\left (x^{2}+\dfrac{1}{x^{2}}\right)$

Again applying same identity:

$x^{4}-\dfrac{1}{x^{4}}$

Hence, option C is correct.

Find the missing term in the following problem.
$\left (\dfrac {3x}{4} - \dfrac {4y}{3}\right )^{2} = \dfrac {9x^{2}}{16} + \dfrac {16y^{2}}{9} + ?$.

  1. $2xy$

  2. $-2xy$

  3. $12xy$

  4. $-12xy$


Correct Option: B
Explanation:
We know that $(a-b)^2=a^2+b^2-2ab$
$\left (\dfrac{3x}{4}-\dfrac{4y}{3}\right)^2=\left (\dfrac{3x}{4}\right)^2+\left (\dfrac{4y}{3}\right)^2-2\left(\dfrac{3x}{4}\right)\left (\dfrac{4y}{3}\right)$
$=\dfrac{9x^2}{16}+\dfrac{16y^2}{9}-2xy$
Therefore, the missing term is $-2xy$.

Option (B) is correct

$(9p - 5q)^{2} + 180 pq$ is equivalent to _______.

  1. $(5p + 9q)^{2}$

  2. $(5p - 9q)^{2}$

  3. $(9p + 5q)^{2}$

  4. $(9p - 5q)^{2}$


Correct Option: C
Explanation:
We know that $a^2+b^2+2ab$
$(a-b)^2=a^2+b^2-2ab$
$\Rightarrow (a-b)^2+4ab=a^2+b^2-2ab+4ab=(a+b)^2$    $(\because a=9p$ and $b=5q)$
$\Rightarrow 4ab=4(9p)(5q)=180pq$
Therefore, $ (9p-5q)^2+180pq=(9p+5q)^2$

Option (C) is correct.

If $\sqrt{\left(12+\sqrt{12+\sqrt{12+....}}\right)}=x$, then the value of x is ____________.

  1. $3$

  2. $4$

  3. $6$

  4. Greater than $6$


Correct Option: B
Explanation:

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

$\sqrt{12+x}=x$
$(12+x)=x^2$
$x^2-x-12=0$
$(x-4)(x+3)=0$
$ x=4$ or $x=-3$
since x is a positive number (eq 1)
$x=4$

$\sqrt { 3+2\sqrt { 2 }  } +\sqrt { 3-2\sqrt { 2 }  } =...$ ?

  1. $2+2\sqrt {2}$

  2. $2\sqrt {2}$

  3. $1$

  4. $0$


Correct Option: B
Explanation:

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


$\Rightarrow$  $\left(\sqrt { 3+2\sqrt { 2 }  } +\sqrt { 3-2\sqrt { 2 }  }\right)^2$          [ Squaring both  sides ]

$\Rightarrow$  $(\sqrt{3+2\sqrt{2}})^2+(\sqrt{3-2\sqrt{2}})^2+2(\sqrt{3+2\sqrt{2}})(\sqrt{3-2\sqrt{2}})$

$\Rightarrow$  $3+2\sqrt{2}+3-2\sqrt{2}+2\sqrt{(3)^2-(2\sqrt{2})^2}$

$\Rightarrow$  $6+2\sqrt{9-8}$

$\Rightarrow$  $8$
Taking square root
$\Rightarrow$  $2\sqrt{2}$

if $y = 500{e^{7x}} + 600{e^{-7x}}$ . Then $\dfrac{{d^2y}}{dx^2} = 49y$

  1. True

  2. False


Correct Option: A
Explanation:

$y=500e^{7x}+600e^{-7x}$


Differentiationg w.r.t. $x$

$\Rightarrow$  $\dfrac{dy}{dx}=500\dfrac{d(e^{7x})}{dx}+600\dfrac{d(e^{-7x})}{dx}$

$\Rightarrow$  $500\times e^{7x}\times \dfrac{d(7x)}{dx}+600\times e^{-7x}\times \dfrac{d(-7x)}{dx}$

$\Rightarrow$  $\dfrac{dy}{dx}=500\times e^{7x}\times 7+600\times e^{-7x}\times (-7)$

$\Rightarrow$  $\dfrac{dy}{dx}=500\times 7\times e^{7x}-600\times 7\times e^{-7x}$

Again differentiating w.r.t. $x$,

$\Rightarrow$  $\dfrac{d^2y}{dx^2}=500\times 7\times \dfrac{d(e^{7x})}{dx}-600\times 7\times \dfrac{d(e^{-7x})}{dx}$

$\Rightarrow$  $\dfrac{d^2y}{dx^2}=500\times 7\times e^{7x}\times 7-600\times 7\times (-7)\times e^{-7x}$

$\Rightarrow$  $\dfrac{d^2y}{dx^2}=500\times 7\times 7e^{7x}+600\times 7\times 7\times e^{-7x}$

$\Rightarrow$  $\dfrac{d^2y}{dx^2}=7\times 7(500e^{7x}+600e^{-7x})$

$\Rightarrow$  $\dfrac{d^2y}{dx^x}=49y$

Evaluate each of the following using identities :
i) $(399)^2$
ii) $(0.98)^2$
iii) $991 \times 1009$

  1. i) 159876

    ii) 0.91
    iii) 876590

  2. i) 135879

    ii) 0.87
    iii) 896750

  3. i) 159201

    ii) 0.9604
    iii) 999919

  4. i) 138760

    ii) 0.9
    iii) 999999


Correct Option: C
Explanation:

$(i)$

$(399)^2=(400-1)^2$
             $=(400)^2+(1)^2-2\times 400\times 1$                [ $(a-b)=a^2+b^2-2ab$ ]
             $=160000+1-800$
             $=159201$
$\therefore$  $(399)^2=159201$

$(ii)$
$(0.98)^2=(1-0.02)^2$
              $=(1)^2+(0.02)^2-2\times 1\times 0.02$                 [ $(a-b)=a^2+b^2-2ab$ ]
              $=1+0.0004-0.04$
              $=0.9604$

$(iii)$ 
$991\times 1009=(1000-9)(1000+9)$
                       $=(1000)^2-(9)^2$                               [ $(a+b)(a-b)=a^2-b^2$ ]     
                       $=1000000-81$ 
                       $=999919$

If $a + b + c = 9$ and $ab + bc + ca = 26$, then the value of $a^2 + b^2 + c^2$ is

  1. $29$

  2. $52$

  3. $81$

  4. None


Correct Option: A
Explanation:

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


Now we have,


$\Rightarrow$$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)$

$\Rightarrow$$a^2+b^2+c^2=(9)^2-2\times26=81-52=29$.

If $x^{2}+2(a-1)x+a+5=0$ has real roots to the interval $(1,3)$, then complete set of value of $'a'$ is

  1. $\left(-\infty,-\dfrac {8}{7}\right)$

  2. $(4,\infty)$

  3. $\left(-\infty,-\dfrac {48}{3}\right)$

  4. $a\ \epsilon \left(-\dfrac {8}{7},-1 \right]$


Correct Option: D
Explanation:

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

                     $\implies (a-1)^{2}-(a+5)\geq 0\implies a\in (-\infty,-1]\cup[4,\infty)\cdots\cdots(1)$
 the roots lies in $(1,3)\implies f(1).f(3)> 0\implies (3{a}+4)(7{a}+8)>0$
                              $\implies a\in (-\infty,-\dfrac{4}{3})\cup(-\dfrac{8}{7},\infty)\cdots\cdots(2)$
from $(1),(2)$    $a\in\bigg(-\dfrac{8}{7},-1\bigg]$

Find the square of $83$ without actual multiplication.

  1. $6880$

  2. $3881$

  3. $3889$

  4. $6889$


Correct Option: D
Explanation:

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

If $x^{2}+\dfrac{1}{^{x^2}}=18$, then the value of $\left(x+\dfrac{1}{x}\right)$ is ?

  1. $1$

  2. $3$

  3. $\sqrt {20}$

  4. $6$


Correct Option: C
Explanation:

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


$\Rightarrow$  $x^2+\dfrac{1}{x^2}=20-2$                           [ Since, $20-2=18$ ]

$\Rightarrow$  $x^2+\dfrac{1}{x^2}+2=20$

$\Rightarrow$  $x^2+\dfrac{1}{x^2}+2\times x\times\dfrac{1}{x}=20$

$\Rightarrow$  $\left(x+\dfrac{1}{x}\right)^2=20$                      [ Since, $a^2+b^2+2ab=(a+b)^2$ ]

$\Rightarrow$  $\left(x+\dfrac{1}{x}\right)=\sqrt{20}$

If $5a\sqrt{b}-\dfrac{3}{2b\sqrt{a}}=12$ and $a=8b$, then the value of $25a^{2}b+\dfrac{9}{4ab^{2}}$ is ?

  1. $144-15\sqrt{8}$

  2. $144+15\sqrt{8}$

  3. $15\sqrt{8}-144$

  4. $-15\sqrt{8}-144$


Correct Option: B
Explanation:

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,


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


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


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


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


$=144+15\sqrt{8}$


If $x+\cfrac{1}{x}=5$, then ${x}^{2}+\cfrac{1}{{x}^{2}}$ is equal to

  1. $25$

  2. $10$

  3. $23$

  4. $27$


Correct Option: C
Explanation:
Given,

$x+\dfrac {1}{x}=5$

squaring on both sides, we get,

$x^2+\dfrac {1}{x^2}+2=25$

$\therefore x^2+\dfrac {1}{x^2}=23$

If ${x}^{2}+\cfrac{1}{{x}^{2}}=102$, then $x-\cfrac{1}{x}$ is equal to

  1. $8$

  2. $10$

  3. $12$

  4. $13$


Correct Option: B
Explanation:
Given,

$\Rightarrow x^2+\dfrac{1}{x^2}=102$

$\Rightarrow \left ( x-\dfrac{1}{x} \right )^2=x^2+\dfrac{1}{x^2}-2$

$\Rightarrow \left ( x-\dfrac{1}{x} \right )^2=102-2=100$

$\Rightarrow x-\dfrac{1}{x} =10$

If ${x}^{3}+\cfrac{1}{{x}^{3}}=110$, then $x+\cfrac{1}{x}$ is equal to

  1. $5$

  2. $10$

  3. $15$

  4. None of these


Correct Option: A
Explanation:

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


$\Rightarrow$  $\left(x+\dfrac{1}{x}\right)^3=110+3\left(x+\dfrac{1}{x}\right)$

Let $y=x+\dfrac{1}{x}$

$\Rightarrow$  $y^3=110+3y$

$\Rightarrow$  $y^3-3y-110=0$

$\Rightarrow$  $y^3-5y^2+5y^2-3y-110=0$

$\Rightarrow$  $y^2(y-5)+5y^2-25y+22y-110=0$

$\Rightarrow$  $y^2(y-5)+5y(y-5)+22(y-5)=0$

$\Rightarrow$  $(y-5)(y^2+5y+22)=0$

$\Rightarrow$  $y=5$ or $y=\dfrac{-5\pm3\sqrt{7}i}{2}$

$\therefore$  $x+\dfrac{1}{x}=5$

If $x+\cfrac{1}{x}=4$, then ${x}^{4}+\cfrac{1}{{x}^{4}}$ is equal to

  1. $196$

  2. $194$

  3. $192$

  4. $190$


Correct Option: B
Explanation:
Given,

$x+\dfrac{1}{x}=4$

squaring on both sides, we get,

$x^2+\dfrac{1}{x^2}+2=16$

$x^2+\dfrac{1}{x^2}=16-2=14$

squaring on both sides, we get,

$x^4+\dfrac{1}{x^4}+2=196$

$x^4+\dfrac{1}{x^4}=196-2=194$

Evaluate $\displaystyle \left ( \frac{2x}{7} - \frac{7y}{4} \right )^{2}$

  1. $\displaystyle \frac{x^{2}}{49} + \frac{17y^{2}}{16} - xy$

  2. $\displaystyle \frac{4x^{2}}{49} + \frac{49y^{2}}{16} - xy$

  3. $\displaystyle \frac{4x^{2}}{9} + \frac{49y^{2}}{4} - xy$

  4. $\displaystyle \frac{x^{2}}{13} + \frac{49y^{2}}{13} - xy$


Correct Option: B
Explanation:

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


Using,

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

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

$=\dfrac{4x^2}{49}-xy+\dfrac{49y^2}{16}$

Evaluate $\displaystyle \left ( \frac{7}{8}x + \frac{4}{5}y \right ) ^{2}$

  1. $\displaystyle \frac{49}{64}x^{2} + \frac{16}{25}y^{3} + \frac{7}{5}xy$

  2. $\displaystyle \frac{78}{32}x^{2} + \frac{16}{25}y^{2} + \frac{1}{5}xy$

  3. $\displaystyle \frac{49}{64}x^{2} + \frac{16}{25}y^{2} + \frac{7}{5}xy$

  4. $\displaystyle \frac{78}{32}x^{2} + \frac{16}{25}y^{2} + \frac{1}{6}xy$


Correct Option: C
Explanation:

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

Find square of the following expression

$\displaystyle 3a - 4b$

  1. $\displaystyle a^{2} - 24ab + 48b^{2}$

  2. $\displaystyle 9a^{2} - 4ab + 48b^{2}$

  3. $\displaystyle 9a^{2} - 24ab + 16b^{2}$

  4. $\displaystyle a^{2} - 24ab + 16b^{2}$


Correct Option: C
Explanation:

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

Find square of the following expression

$\displaystyle \frac{3a}{2b} - \dfrac{2b}{3a}$

  1. $\displaystyle \dfrac{9a^{2}}{4b^{2}} - 2 + \dfrac{b^{2}}{a^{2}}$

  2. $\displaystyle \dfrac{9a^{2}}{4b^{2}} + 2 + \dfrac{4b^{2}}{9a^{2}}$

  3. $\displaystyle \dfrac{a^{2}}{4b^{2}} - 2 + \dfrac{4b^{2}}{9a^{2}}$

  4. $\displaystyle \dfrac{9a^{2}}{4b^{2}} - 2 + \dfrac{4b^{2}}{9a^{2}}$


Correct Option: D
Explanation:

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

Find the square of  $\displaystyle a + 2b + c$

  1. $\displaystyle a^{2} + b^{2} + c^{2} + ab + bc + ac$

  2. $\displaystyle a^{2} + 4b^{2} + c^{2} + 4ab + 4bc + 2ac$

  3. $\displaystyle a^{3} + 4b^{3} + c^{3} + 8ab + 8bc + 8ac$

  4. $\displaystyle a^{3} + b^{3} + c^{3} + 4ab + 4bc + 2ac$


Correct Option: B
Explanation:

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

Find the square of $\displaystyle 2a - b - 3c$

  1. $\displaystyle 4a^{2} + b^{2} + 9c^{2} - 4ab + 6bc - 12ca$

  2. $\displaystyle a^{3} + b^{3} + c^{2} - 4ab + 6bc - ca$

  3. $\displaystyle a^{2} + b^{2} - c^{2} - 4ab + 6bc - 2ca$

  4. $\displaystyle 4a^{3} + b^{3} + 9c^{2} - 4ab - 6bc - 12ca$


Correct Option: A
Explanation:

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

Evaluate :

$\displaystyle \left ( \frac{2x}{5} + \frac{3y}{4} - \frac{4z}{7} \right )^{2}$

  1. $\displaystyle \frac{x^{2}}{25} + \frac{9y^{2}}{16} + \frac{z^{2}}{49} + \frac{3}{5} xy - \frac{6}{7}yz - \frac{16}{35}zx$

  2. $\displaystyle \frac{4x^{2}}{25} + \frac{9y^{2}}{16} + \frac{16z^{2}}{49} + \frac{1}{5} xy - \frac{6}{3}yz - \frac{36}{35}zx$

  3. $\displaystyle \frac{x^{2}}{25} + \frac{9y^{2}}{16} + \frac{5z^{2}}{49} + \frac{3}{5} xy - \frac{6}{7}yz - \frac{16}{35}zx$

  4. $\displaystyle \frac{4x^{2}}{25} + \frac{9y^{2}}{16} + \frac{16z^{2}}{49} + \frac{3}{5} xy - \frac{6}{7}yz - \frac{16}{35}zx$


Correct Option: D
Explanation:

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

Find the squares of the following numbers without actual multiplication:
$49$
$52$

  1. $2441; 2704$

  2. $2401; 2784$

  3. $2401; 2704$

  4. $2441; 2784$


Correct Option: C
Explanation:

$49^2=(40+9)^2=40(40+9)+9(40+9)$
$\;\;\;\;\;=(40)^2+40\times9+9\times40+9^2$
$\;\;\;\;\;=1600+360+360+81$
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, $52^2=2704$.

Find the square of $43$ without multiplication.

  1. $2401$

  2. $4801$

  3. $1842$

  4. $1849$


Correct Option: D
Explanation:

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.

Find the square of $125$.

  1. $84113$

  2. $48000$

  3. $15625$

  4. $84920$


Correct Option: C
Explanation:

The square of $125^2$

$ = (120+5)^2$

$=14400+25+1200$

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

Without doing multiplication, find the square of $29.$

  1. $841$

  2. $480$

  3. $742$

  4. $849$


Correct Option: A
Explanation:

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.

Without actual finding the square of the numbers, find the value of $120^2 - 119^2$.

  1. $239$

  2. $240$

  3. $238$

  4. $237$


Correct Option: A
Explanation:

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

Without actual finding the square of the numbers, find the value of $36^2 - 35^2$.

  1. $70$

  2. $71$

  3. $72$

  4. $73$


Correct Option: B
Explanation:

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

Evaluate the following
$(0.98)^{2}$

  1. $0.9664$

  2. $0.9604$

  3. $0.9864$

  4. $0.9964$


Correct Option: B
Explanation:

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


${ (0.98) }^{ 2 }\ =(1-0.02)^{ 2 }\ =1^{ 2 }+(0.02)^{ 2 }-(2\times 1\times 0.02)\quad \quad \quad \quad \quad (\because \quad (a-b)^{ 2 }=a^{ 2 }+b^{ 2 }-2ab)\ =1+0.0004-0.04\ =1.0004-0.04\ =0.9604$

Hence, ${ (0.98) }^{ 2 }=0.9604$

A factor of $(3x^{4} - 12y^{4})$ is _________.

  1. $3$

  2. $x^{2} - 2y^{2}$

  3. $x^{2} + 2y^{2}$

  4. All of these


Correct Option: A
Explanation:
Given expression is $(3x^{4}-12y^{4})$
Taking $3$ common, we get
$= 3(x^{4}-4y^{4})$
$= 3(x^4-(\sqrt{2}y)^{4})$
We know the identity and applying it $a^{2}-b^{2} = (a-b)(a+b)$

$3x^{4}-12y^{4}= 3(x^{4}-(\sqrt{2}y)^{4})$

$3x^{2}-12y^{2}= 3(x^{2}-\sqrt{2}y^{2})(x^{2}+\sqrt{2}y^{2})$

Find the square of the following number without multiplication.

46

  1. 2116

  2. 2002

  3. 2424

  4. 1988


Correct Option: A
Explanation:

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

If sin$\theta -cosec  \theta =\sqrt{5},$ then the value of sin  $\theta  + cosec  \theta$ is:

  1. $\sqrt{3}$

  2. 1

  3. 3

  4. 9


Correct Option: C
Explanation:

$\Rightarrow \sin\theta-cosec\theta=\sqrt{5}$


$\Rightarrow \sin\theta-\dfrac{1}{\sin\theta}=\sqrt{5}$      $(\because cosec\theta=\dfrac{1}{\sin\theta})$

$\Rightarrow \sin^2\theta-\sqrt{5}\sin\theta-1=0$

Solving equation to get roots.

$\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{\pm3+\sqrt{5}}{2}$ (substitute values to get roots)

To find:-

$\sin\theta+cosec\theta$

Using $\dfrac{3+\sqrt{5}}{2}$

$=\dfrac{3+\sqrt{5}}{2}+\dfrac{2}{3+\sqrt{5}}$

$=\dfrac{(3+\sqrt{5})^2+4}{2(3+\sqrt{5})}$

$=\dfrac{9+4+5+6\sqrt{5}}{2(3+\sqrt{5})}$

$=\dfrac{6(3+\sqrt{5})}{2(3+\sqrt{5})}$

$=3$


Using $\dfrac{-3+\sqrt{5}}{2}$

$=\dfrac{-3+\sqrt{5}}{2}+\dfrac{2}{-3+\sqrt{5}}$

$=\dfrac{(-3+\sqrt{5})^2+4}{2(-3+\sqrt{5})}$

$=\dfrac{9+4+5-6\sqrt{5}}{2(-3+\sqrt{5})}$

$=\dfrac{-6(-3+\sqrt{5})}{2(-3+\sqrt{5})}$

$=-3$


According to option answer is $3$

If $a+b+c=6$ and $ ab+bc+ca = 11 $
Find $\left( { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } \right)$ ?
  1. $14$

  2. $25$

  3. $36$

  4. $47$


Correct Option: A
Explanation:
Given $a+b+c=6$ 

$(a+b+c)^2=36$

$(a^2+b^2+c^2)+2(ab+ba+ca)=36$

$a^2+b^2+c^2+(2\ast 11)=36$

$a^2+b^2+c^2=36-22=14$

Evaluating the following :
$(3+\sqrt{2})^{5}-(3-\sqrt{2})^{5}$

  1. $1718\sqrt 3$

  2. $1718\sqrt 2$

  3. $1178\sqrt 3$

  4. $1178\sqrt 2$


Correct Option: D
Explanation:

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


$\Rightarrow 2\left[\ ^{5}C _{1}\times 3^{4}\times (\sqrt{2})^{1}+\ ^{5}C _{3}\times 3^{2}\times (\sqrt{2})^{3}+\ ^{5}C _{5}\times 3^{0}\times (\sqrt{2})^{5}\right]$

$\Rightarrow 2\left[5\times 81\times\sqrt{2}+10\times 9\times 2\sqrt{2}+4\sqrt{2}\right]$

$\Rightarrow 2\sqrt{2}(405+180+4)$

$\Rightarrow 1178\sqrt{2}$

Evaluating the following :
$(1+2\sqrt{x})^{5}+(1-2\sqrt{x})^{5}$

  1. $2(1+40x^2+80x)$

  2. $2(1-40x+81x^2)$

  3. $2(1+40x+80x^2)$

  4. None of these


Correct Option: C
Explanation:
Given to evaluate is $(1+2\sqrt x)^5 +(1-2\sqrt x)^5$

$\Rightarrow 2[^{5}C _{0} (2\sqrt x)^0 +^5C _2 (2\sqrt x)^2 +^5C _4 (2\sqrt x)^4]$

$\Rightarrow 2[1+10\times 4x+5\times 16x^2]$

$\Rightarrow 2[1+40x+80x^2]$

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