Questions Related to maths

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

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

  1. $2$
  2. $4$
  3. $6$
  4. $8$
Reveal answer Fill a bubble to check yourself
D Correct answer
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
$\dfrac{x^{2} + \dfrac{1}{x^{2}} - 2x\dfrac{1}{x}}{\sqrt{6}} = \sqrt{6}$
$\Rightarrow x^{2} + \dfrac{1}{x^{2}} - 2  =\sqrt{6}\times \sqrt{6}$
$\Rightarrow x^{2} + \dfrac{1}{x^{2}} = 6+2 = 8$

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

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

  1. $239$
  2. $240$
  3. $241$
  4. $361$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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. $x^{4} + \dfrac {1}{x^{4}}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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. $-12xy$
Reveal answer Fill a bubble to check yourself
B Correct answer
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
Multiple choice maths squares and square roots finding the square of a number finding square of a number patterns in square numbers

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

  1. $(5p + 9q)^{2}$
  2. $(5p - 9q)^{2}$
  3. $(9p + 5q)^{2}$
  4. $(9p - 5q)^{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.