Tag: estimating square roots
Questions Related to estimating square roots
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$46$
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$50$
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$52$
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$58$
Its unit digit is 6 and tens digit is 1
The number which exceeds its positive square root by $12$ is
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$9$
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$16$
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$25$
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None of these
Find the square root of which of the following numbers will be the least :
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$7\dfrac{58}{81}$
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$11\dfrac{14}{25}$
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$10\dfrac{1}{36}$
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$0.3481$
$A.$
Simlify: $\sqrt{\dfrac{-17}{144}-i}$
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$ \pm \left( {\dfrac{3}{2} - \dfrac{i}{3}} \right)$
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$ \pm \left( {\dfrac{3}{4} - \dfrac{{2i}}{3}} \right)$
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$ \pm \left( {\dfrac{3}{5} - \dfrac{{5i}}{6}} \right)$
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$ \pm \left( {\dfrac{2}{3} - \dfrac{{3i}}{4}} \right)$
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.
The approximate value of$\sqrt { { \left( 1.97 \right) }^{ 2 }{ \left( 4.02 \right) }^{ 2 }{ \left( 3.98 \right) }^{ 2 } }$
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$31.59 $
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$5.099$
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$5.009$
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$5.734$
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.
Find the square root correct upto $2$ decimals $60.92$.
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$7.92$
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$7.81$
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$7.27$
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$7.56$
The positive square root of $( \sqrt { 48 } - \sqrt { 45 } )$ is _________.
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$\frac { \sqrt [ 4 ] { 3 } } { \sqrt { 2 } } ( \sqrt { 5 } - \sqrt { 3 } )$
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$\frac { \sqrt [ 4 ] { 3 } } { 2 } ( \sqrt { 5 } - \sqrt { 3 } )$
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$\frac { \sqrt { 2 } } { \sqrt [ 4 ] { 3 } } ( \sqrt { 5 } - \sqrt { 3 } )$
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$\frac { \sqrt [ 4 ] { 3 } } { \sqrt { 2 } } ( \sqrt { 5 } + \sqrt { 3 } )$
Solving $(\sqrt{48}-\sqrt{45})$
If $\sqrt{2}=1.414$ then the value of $\sqrt{8}$ is
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$2.828$
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$1.828$
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$2.282$
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$2.288$
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$\sqrt{13}-\sqrt{2}$
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$\sqrt{3}-\sqrt{2}$
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$\sqrt{5}-\sqrt{3}$
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$\sqrt{5}-\sqrt{2}$
$5-2\sqrt{6}=3+2-2\sqrt{6}$
$=({\sqrt{3}})^2+({\sqrt{2}})^2-2\sqrt{3}\times \sqrt {2}$
${5-2\sqrt{6}}=(\sqrt{3}-\sqrt{2})^2$
$\sqrt {5-2\sqrt{6}}=(\sqrt{3}-\sqrt{2})$
$So, \ the \ square \ root \ of \ (5-2\sqrt{6})=\sqrt{3}-\sqrt{2}$
$\sqrt{(a - b)^2} + \sqrt{(b - a)^2}$ is
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Always zero
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Never zero
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Positive if and only if a > b
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Positive only if a $\ne$ b
$\sqrt{(a - b)^2} + \sqrt{(b - a)^2}$
$= |a - b| + |b - a|$
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.