Tag: finding square of a number

Questions Related to finding square of a number

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 ?

maths squares and square roots finding the square of a number finding square of a number patterns in square numbers

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

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

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

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

Show answer
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

maths squares and square roots finding the square of a number finding square of a number patterns in square numbers

  1. $25$

  2. $10$

  3. $23$

  4. $27$

Show answer
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

maths squares and square roots finding the square of a number finding square of a number patterns in square numbers

  1. $8$

  2. $10$

  3. $12$

  4. $13$

Show answer
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

maths squares and square roots finding the square of a number finding square of a number patterns in square numbers

  1. $5$

  2. $10$

  3. $15$

  4. None of these

Show answer
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

maths squares and square roots finding the square of a number finding square of a number patterns in square numbers

  1. $196$

  2. $194$

  3. $192$

  4. $190$

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

maths squares and square roots finding the square of a number finding square of a number patterns in square numbers

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

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

maths squares and square roots finding the square of a number finding square of a number patterns in square numbers

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

Show answer
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

$\displaystyle 3a - 4b$

maths squares and square roots finding the square of a number finding square of a number patterns in square numbers

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

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

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

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

Show answer
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

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

maths squares and square roots finding the square of a number finding square of a number patterns in square numbers

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

Show answer
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  $\displaystyle a + 2b + c$

maths squares and square roots finding the square of a number finding square of a number patterns in square numbers

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

Show answer
Correct Option: B
Explanation:

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