Tag: square root of non perfect squares

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Multiple choice maths squares and square roots approximation of square roots estimating square roots square root of non perfect squares

The number which exceeds its positive square root by $12$ is

  1. $9$
  2. $16$
  3. $25$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Let the positive number be x according to question,

$\sqrt{x}+12=x$

$\Rightarrow \sqrt{x}=x-12$

Squaring both sides,

$\Rightarrow x=x^{2}-24x+144$

$x^{2}-25x+144=0$

$x^{2}-16x-9x+144=0$

$x(x-16)-9(x-16)=0$

$(x-9)(x-16)=0$

 $  (x-9)=0 $ or $    (x-16)=0 $

 $ x=9 $  or $ x=16 $

So, the number is $16$.
Multiple choice maths squares and square roots approximation of square roots estimating square roots square root of non perfect squares

Find the square root of which of the following numbers will be the least :

  1. $7\dfrac{58}{81}$
  2. $11\dfrac{14}{25}$
  3. $10\dfrac{1}{36}$
  4. $0.3481$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$A.$


$7\dfrac{58}{81}=\dfrac{625}{81}$


$\Rightarrow$  $\sqrt{\dfrac{625}{81}}=\dfrac{25}{9}=2.77$

$B.$

$11\dfrac{14}{25}=\dfrac{289}{25}$

$\Rightarrow$  $\sqrt{\dfrac{289}{25}}=\dfrac{17}{5}=3.4$

$C.$

$10\dfrac{1}{36}=\dfrac{361}{36}$

$\Rightarrow$  $\sqrt{\dfrac{361}{36}}=\dfrac{19}{6}=3.16$

$D.$

$0.3481=\dfrac{3481}{10000}$

$\Rightarrow$  $\sqrt{\dfrac{3481}{10000}}=\dfrac{59}{100}=0.59$

$\therefore$  We can see, $0.3481$  has least square root.

Multiple choice maths squares and square roots approximation of square roots estimating square roots square root of non perfect squares

Simlify: $\sqrt{\dfrac{-17}{144}-i}$

  1. $ \pm \left( {\dfrac{3}{2} - \dfrac{i}{3}} \right)$
  2. $ \pm \left( {\dfrac{3}{4} - \dfrac{{2i}}{3}} \right)$
  3. $ \pm \left( {\dfrac{3}{5} - \dfrac{{5i}}{6}} \right)$
  4. $ \pm \left( {\dfrac{2}{3} - \dfrac{{3i}}{4}} \right)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We know that,

${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$

 

Now, let,

$ -2ab=-i $

$ ab=i $

 

Now, consider $\dfrac{-17}{144}$. We can write it as,

$\dfrac{-17}{144}=\dfrac{64-81}{9\times 16}=\dfrac{4}{9}-\dfrac{9}{16}$

 

Thus,

$ \sqrt{\dfrac{-17}{144}-i}=\sqrt{\dfrac{4}{9}-\dfrac{9}{16}-i} $

$ =\sqrt{{{\left( \dfrac{2}{3} \right)}^{2}}+{{\left( i \right)}^{2}}{{\left( \dfrac{3}{4} \right)}^{2}}-2\times \left( \dfrac{2}{3} \right)\times \left( \dfrac{3i}{4} \right)} $

$ =\sqrt{{{\left( \dfrac{2}{3}-\dfrac{3i}{4} \right)}^{2}}} $

$ =\pm \left( \dfrac{2}{3}-\dfrac{3i}{4} \right) $

 

Hence, this is the required result.

Multiple choice maths squares and square roots approximation of square roots estimating square roots square root of non perfect squares

The approximate value of$\sqrt { { \left( 1.97 \right)  }^{ 2 }{ \left( 4.02 \right)  }^{ 2 }{ \left( 3.98 \right)  }^{ 2 } }$

  1. $31.59 $
  2. $5.099$
  3. $5.009$
  4. $5.734$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We have,

$ \sqrt{{{\left( 1.97 \right)}^{2}}{{\left( 4.02 \right)}^{2}}{{\left( 3.98 \right)}^{2}}} $

$ =1.97\times 4.02\times 3.98 $

$ =31.59 $

Hence, this is the answer.

Multiple choice maths squares and square roots approximation of square roots estimating square roots square root of non perfect squares

The positive square root of $( \sqrt { 48 } - \sqrt { 45 } )$ is _________.

  1. $\frac { \sqrt [ 4 ] { 3 } } { \sqrt { 2 } } ( \sqrt { 5 } - \sqrt { 3 } )$
  2. $\frac { \sqrt [ 4 ] { 3 } } { 2 } ( \sqrt { 5 } - \sqrt { 3 } )$
  3. $\frac { \sqrt { 2 } } { \sqrt [ 4 ] { 3 } } ( \sqrt { 5 } - \sqrt { 3 } )$
  4. $\frac { \sqrt [ 4 ] { 3 } } { \sqrt { 2 } } ( \sqrt { 5 } + \sqrt { 3 } )$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Solving $(\sqrt{48}-\sqrt{45})$

$4\sqrt{3}-3\sqrt{5}$

$\dfrac{\sqrt{3}}{2}(8-2\sqrt{15})$

$\dfrac{\sqrt{3}}{2}(3+5-2\sqrt{15})$

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

$\dfrac{\sqrt3}{2}(\sqrt5-\sqrt3)^2$

$Now \ Finding\ Square \ Root$

$\pm{ \dfrac{3^{\frac{1}{4}}}{\sqrt2}(\sqrt5-\sqrt3)}$

$So \ it's \ positive \ root \ is \ $$ \dfrac{3^{\frac{1}{4}}}{\sqrt2}(\sqrt5-\sqrt3)$
Correct Answer is $A$

Multiple choice maths squares and square roots approximation of square roots estimating square roots square root of non perfect squares

Find the square root of 
$5-2\sqrt{6}$

  1. $\sqrt{13}-\sqrt{2}$
  2. $\sqrt{3}-\sqrt{2}$
  3. $\sqrt{5}-\sqrt{3}$
  4. $\sqrt{5}-\sqrt{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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


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

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


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

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

$So, \  the \  square \  root \  of \  (5-2\sqrt{6})=\sqrt{3}-\sqrt{2}$

Multiple choice maths squares and square roots approximation of square roots estimating square roots square root of non perfect squares

$\sqrt{(a - b)^2} + \sqrt{(b - a)^2}$ is

  1. Always zero

  2. Never zero

  3. Positive if and only if a > b
  4. Positive only if a $\ne$ b
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\sqrt{(a - b)^2} + \sqrt{(b - a)^2}$
$= |a - b| + |b - a|$

Now, If $a > b$
$= a - b + a - b$
$= 2a - 2b$...+ ve

If $b > a$
$= b - a + b - a$
$= 2b - 2a$...+ ve
Therefore, if $a \ne b$ then the given equation is always positive.
Hence, option 'D' is correct.