Question

which of the following is an not perfect square. a)361. b)3481. c)4624. d)8024.​


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Answer 1

Step-by-step explanation:

a) 361= 19 × 19 is a perfect square number

b) 3481= 59×59 is a perfect square number

c) 4624= 68×68 is a perfect square number

89 × 89 = 7921< 8024< 90 × 90 = 8100

d) 8024 is not a perfect square number

Answer 2

$\bf \underline{\underline{\maltese\: Solution }}$

$\bf (a) \: \: 361$

$\sf First \: of \: all \: we \: have \: to \: find \: prime \: factors \: of \: 361.$

$\begin{gathered}\begin{array}{c | c} \underline{19 }&\underline{361} \\ \underline{19}&\underline{19 }\\ &1\end{array}\end{gathered}$

$\sf \implies 361 = \underbrace{19 \times 19 }$

$\sf Thus, \: 361 \: can \: be \: expressed \: as \: pair \: of \: equal \: factors.$

$\sf \underline{Therefore, \: 361 \: is \: a \: perfect \: square. }$

$\: \: \: \: \: \: \: \: \: \: \: \:$

$\bf (b) \: \: 3481$

$\sf First \: of \: all \: we \: have \: to \: find \: prime \: factors \: of \: 3481.$

$\sf \begin{gathered}\begin{array}{c | c} \underline{59 }&\underline{3481} \\ \underline{59}&\underline{59 }\\ &1\end{array}\end{gathered}$

$\sf \implies 3481 = \underbrace{59 \times 59 }$

$\sf Thus, \: 3481 \: can \: be \: expressed \: as \: pair \: of \: equal \: factors.$

$\sf \underline{Therefore, \: 3481 \: is \: a \: perfect \: square. }$

$\: \: \: \: \: \: \: \: \: \: \: \:$

$\bf (c) \: \: 4624$

$\sf First \: of \: all \: we \: have \: to \: find \: prime \: factors \: of \: 4624.$

$\sf \begin{gathered}\begin{array}{c | c} \underline{2}&\underline{4624} \\\underline{2}&\underline{2312} \\ \underline{2}&\underline{1156} \\ \underline{2}&\underline{578} \\ \underline{17}&\underline{289 }\\\underline{17}&\underline{17} \\ &1\end{array}\end{gathered}$

$\sf \implies 4624 = \underbrace{2 \times 2 } \times \underbrace{2 \times 2} \times \underbrace{17 \times 17 }$

$\sf Thus, \: 4624 \: can \: be \: expressed \: as \: pair \: of \: equal \: factors.$

$\sf \underline{Therefore, \: 4624 \: is \: a \: perfect \: square. }$

$\: \: \: \: \: \: \: \: \: \: \: \:$

$\bf (d) \: \: 8024$

$\sf First \: of \: all \: we \: have \: to \: find \: prime \: factors \: of \: 8024.$

$\sf \begin{gathered}\begin{array}{c | c} \underline{2}&\underline{8024} \\\underline{2}&\underline{4012} \\ \underline{2}&\underline{2006 }\\\underline{1003}&\underline{1003} \\ &1\end{array}\end{gathered}$

$\sf \implies 8024 = \underbrace{2 \times 2 } \times 2 \times 1003$

$\sf \: Making \: pairs \: of \: equal \: factors, \: we \: find \: 2 \: and \: 1003 \: left.$

$\sf \underline{Therefore, \: 8024 \: is \:not \: a \: perfect \: square. }$