Question

1. Find by prime factorization the square root of the following numbers:
(i) 2916 (ii) 2704 (iii) 7056
(iv) 9025​

Answer 1

Step-by-step explanation:

hope it will helps you

Answer 2

$\sf \: \underline{To \: find \: the \: square \: root \: of \: a \: perfect \: square \: by \: prime \: factorization}$

$\sf \: \underline{We \: have \: to \: follow \: the \: steps \: given \: below} :$

$\bf{[Step : 1]}$

$\small\sf{Express \: the \: given \: number \: as \: a \: product \: of \: prime \: factors.}$

$\bf{[Step : 2] }$

$\small \sf{Arrange \: the \: factors \: and \: group \: them \: in \: pairs.}$

$\bf [Step : 3]$

$\small \sf \: Choose \: one \: prime \: factor \: from \: each \: pair \: and \: multiply \: all \: such \: primes.$

$\bf [Step : 4]$

$\small \sf \: The \: product \: is \: the \: square \: root \: of \: given \: number.$

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Solution : 1

$\sf \: We \: have \: to \: find \: the \: prime \: factor \: of \: 2916$

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{2}}}&{\underline{\sf{\:\:2916\:\:\:}}}\\ {\underline{\sf{2}}}&{\underline{\sf{\:\:1458\:\:\:}}}\\ {\underline{\sf{3}}}&{\underline{\sf{\:\:729\:\:\:}}}\\ {\underline{\sf{3}}}&{\underline{\sf{\:\:243\:\:\:}}}\\ {\underline{\sf{3}}}&{\underline{\sf{\:\:81\:\:\:}}}\\ {\underline{\sf{3}}}&{\underline{\sf{\:\:27\:\:\:}}}\\ {\underline{\sf{3}}}& \underline{\sf{\:\:9\:\:\:}} \\\underline{\sf{3}}&{\underline{\sf{\:\:3\:\:\:}}} \\ \underline{\sf{}}&{\sf{\:\:1\:\:\:}} \end{array}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$

$\sf 2916 = \underline{2 \times 2} \times \underline{3 \times 3} \times \underline{3 \times 3} \times \underline{3 \times 3}$

$\sf \sqrt{2916} = \sqrt{2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3}$

$\bold{\therefore } \: \sf \sqrt{2916} = \sqrt{2 \times 3 \times3 \times 3 }$

$\sf \sqrt{2916} = 54$

$\underline{\red{ \sf \: Thus, 54 \: is \: the \: square \: root \: of \: 2916.}}$

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Solution : 2

$\sf \: We \: have \: to \: find \: the \: prime \: factor \: of \: 2704$

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered} \:\: \begin{array}{c|c} {\underline{\sf{2}}}&{\underline{\sf{\:\:2704\:\:\:}}}\\ {\underline{\sf{2}}}&{\underline{\sf{\:\:1352\:\:\:}}}\\ {\underline{\sf{2}}}&{\underline{\sf{\:\:676\:\:\:}}}\\ {\underline{\sf{2}}}&{\underline{\sf{\:\:338\:\:\:}}}\\ {\underline{\sf{13}}}&{\underline{\sf{\:\:168\:\:\:}}}\\ {\underline{\sf{13}}}&{\underline{\sf{\:\:13\:\:\:}}} \\ \underline{\sf{}}&{\sf{\:\:1\:\:\:}} \end{array}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$

$\sf 2704 = \underline{2 \times 2} \times \underline{2 \times 2} \times \underline{13 \times 13}$

$\sf \sqrt{2704} = \sqrt{2 \times 2\times 2 \times 2 \times 13 \times 13}$

$\sf \bold{ \therefore} \: \sqrt{2704} = \sqrt{2 \times 2 \times 13}$

$\sf \sqrt{2704} = 52$

$\underline \green{ \sf \: Thus, 52 \: is \: the \: square \: root \: of \: 2704}$

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