Solutions to Simplify Radicals Questions

Solutions with detailed explanations to questions on simplifying radicals are presented.

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    In the solutions below, we use the product rule of radicals given by
    \( \sqrt{x \times y} = \sqrt{x } \sqrt{y} \)


  1. Simplify the expression \( 2 \sqrt{50} + 12 \sqrt{8} \).
    Solution
    Use the fact that \( 50 = 2 \times 25 \) and \( 8 = 2 \times 4 \) to rewrite the given expressions as follows
    \( \quad 2 \sqrt{50} + 12 \sqrt{8} = 2 \sqrt {2 \times 25} + 12 \sqrt{2 \times 4} \)
    Use the product rule of radicals \( \sqrt{x \times y} = \sqrt{x } \sqrt{y} \) to rewrite the above as
    \( \quad = 2 \sqrt 2 \sqrt{25} + 12 \sqrt 2 \sqrt 4 \)
    Use the fact that \( \sqrt{25} = 5 \) and \( \sqrt 4 = 2\) to rewrite the above as
    \( \quad = 2 \times 5 \times \sqrt 2 + 12 \times 2 \times \sqrt 2 \)
    Simplify
    \( \quad = 10 \sqrt 2 + 24 \sqrt 2 \)
    Factor \( \sqrt 2 \) out and simplify
    \( \quad = (10 + 24) \sqrt 2 = 34 \sqrt 2 \)


  2. Simplify the expression \( \sqrt{27} - \sqrt{300} \).
    Solution
    Knowing that \( 27 = 3 \times 9 \) and \( 300 = 3 \times 100 \), we write
    \( \sqrt{27} - \sqrt{300} = \sqrt{3 \times 9} - \sqrt {3 \times 100} \)
    Use the rule for radicals of radicals \( \sqrt{x \times y} = \sqrt{x } \sqrt{y} \) to rewrite the above as
    \( = \sqrt{3} \sqrt{9} - \sqrt {3} \sqrt{100} \)
    Knowing that \( \sqrt{9} = 3 \) and \( \sqrt{100} = 10\), the above may be simplified to
    \( = 3 \sqrt 3 - 10 \sqrt3 = (3 - 10 )\sqrt 3 = - 7 \sqrt 3 \)


  3. Simplify the expression \( - 2 \sqrt{16y} + 10 \sqrt y \).
    Solution
    Use the product rule of radicals \( \sqrt{x \times y} = \sqrt{x } \sqrt{y} \) to rewrite \( \sqrt{16y} \) as \( \sqrt {16} \sqrt y \) and substitute in the given expression
    \( - 2 \sqrt{16y} + 10 \sqrt y = - 2 \sqrt {16} \sqrt y + 10 \sqrt y \)
    Use the fact that \( \sqrt {16} = 4 \) and substitute in the above
    \( = - 2 \times 4 \sqrt y + 10 \sqrt y \)
    \( = - 8 \sqrt y + 10 \sqrt y \)
    Factor \( \sqrt y \) out and simplify
    \( = (-8 + 10) \sqrt y = 2 \sqrt y \)



  4. Simplify the expression \( 2 \sqrt{x + 1} + 3 \sqrt{16x + 16} \).
    Solution
    Factor 16 out in \( 16x + 16 \)
    \( 2 \sqrt(x + 1) + 3 \sqrt(16x + 16) = 2 \sqrt{x + 1} + 3 \sqrt{16(x + 1) } \)
    Use the product rule of radicals to write \( \sqrt{16(x + 1)} \) as \( \sqrt{16} \sqrt {x-1} = 4 \sqrt {x-1} \) and substitute in the above
    \( = 2 \sqrt{x + 1} + 3 \times 4 \times \sqrt{x + 1} \)
    Simplify
    \( = 2 \sqrt{x + 1} + 12 \sqrt{x + 1} \)
    Factor \(\sqrt{x + 1} \) and simplify
    \( = ( 2 + 12 ) \sqrt{x + 1} = 14 \sqrt{x + 1} \)


  5. \( 2 \sqrt 3 + 4 \sqrt{12} + 3 \sqrt{48} = \)
    Solution
    Write the radicands as product of numbers whose square roots are integers: \( 12 = 4 \times 3 \) and \( 48 = 3 \times 16 \)
    \( 2 \sqrt3 + 4 \sqrt{12} + 3 \sqrt{48} = 2 \sqrt3 + 4 \sqrt{3 \times 4} + 3 \sqrt{3 \times 16} \)
    Use the product rule of radicals to write the above as
    \( = 2 \sqrt3 + 4 \sqrt3 \sqrt4 + 3 \sqrt3 \sqrt{16} \)
    Simplify \( \sqrt4 \) and \( \sqrt{16} \)
    \( = 2 \sqrt3 + 8 \sqrt3 + 12 \sqrt3 \)
    Factor \( \sqrt3 \) and rewrite the above as
    \( = (2 + 8 + 12) \sqrt3 \)
    Simplify
    \( = 22 \sqrt3 \)


  6. Rewrite the expression \( \dfrac {\sqrt3 + \sqrt{12}} {\sqrt 3 - \sqrt{12}} \) without radicals.
    Solution
    We rewrite the given expression using the fact that \( \sqrt{12} = \sqrt{4 \times 3} = \sqrt{4} \sqrt{3} = 2 \sqrt 3 \)
    \( \dfrac {\sqrt3 + \sqrt{12}} {\sqrt 3 - \sqrt{12}} = \dfrac {\sqrt3 + 2 \sqrt{3}} {\sqrt 3 - 2 \sqrt{3}} \)
    Group terms in the numerator and denominator
    \( = \dfrac {(1+2) \sqrt{3}} {(1-2) \sqrt{3}} = \dfrac{3 \sqrt 3}{-\sqrt 3} \)
    Divide numerator and denominator by \( \sqrt 3 \) and simplify (cancel)
    \( = - 3 \)


  7. Simplify the expression \( 5 \sqrt x + 6 \sqrt {9x} - 10 \sqrt {16x} \).
    Solution
    Use the product rule of radicals to rewrite the above as
    \( 5 \sqrt x + 6 \sqrt {9x} - 10 \sqrt {16x} = 5 \sqrt x + 6 \sqrt9 \sqrt x - 10 \sqrt{16} \sqrt x \)
    Simplify \( \sqrt 9 \) and \( \sqrt{16} \)
    \( = 5 \sqrt x + 6 \times 3 \sqrt x - 10 \times 4 \sqrt x \)
    Simplify
    \( = 5 \sqrt x + 18 \sqrt x - 40 \sqrt x \)
    Factor \( \sqrt x \) and simplify
    \( = (5 + 18 - 40) \sqrt x = -17 \sqrt x \)


  8. \( 2 \sqrt {27} + 2 \sqrt{75} = \)
    Solution
    Write 27 and 75 as products of numbers whose square roots are integers: \( 27 = 3 \times 9 \) and \( 75 = 3 \times 25 \)
    \( 2 \sqrt {27} + 2 \sqrt{75} = 2 \sqrt {3 \times 9} + 2 \sqrt{3 \times 25} \)
    Use the product rule of radicals
    \( = 2 \sqrt 3 \sqrt 9 + 2 \sqrt{3} \sqrt{ 25} \)
    Simplify using \( \sqrt 9 = 3 \) and \( \sqrt{ 25} = 5 \)
    \( = 6 \sqrt 3 + 10 \sqrt{3} \)
    Factor \( \sqrt{3} \) out and simplify
    \( = (6 + 10) \sqrt{3} = 16 \sqrt{3} \)

  9. \( \sqrt {10^3} + \sqrt {10^5} = \)
    Solution
    Write \( 10^3 \) and \( 10^5 \) as products of numbers whose square roots are integers: \( 10^3 = 10 \times 100 \) and \( 10^5 = 10 \times 10000 \)
    \( \sqrt {10^3} + \sqrt {10^5} = \sqrt {10 \times 100} + \sqrt{10 \times 10000} \)
    Use the product rule of radicals
    \( = \sqrt {10} \sqrt {100} + \sqrt{10} \sqrt{ 10000} \)
    Simplify using \( \sqrt 100 = 10 \) and \( \sqrt{ 10000} = 100 \)
    \( = 10 \sqrt 10 + 100 \sqrt{10} \)
    Factor \( \sqrt{10} \) out and simplify
    \( = (10 + 100) \sqrt{10} = 110 \sqrt{10} \)

  10. Simplify and rewrite the expression \( \sqrt 8 \sqrt 3 \sqrt 6 \) without radicals.
    Solution
    Write \( 8 \) as \( 2 \times 4 \)
    \( \sqrt 8 \sqrt 3 \sqrt 6 = \sqrt{ 4 \times 2 } \sqrt 3 \sqrt 6 \)
    Use the product rule of radicals on \( \sqrt{ 4 \times 2 } \) and simplify
    \( = \sqrt 4 \sqrt 2 \sqrt 3 \sqrt 6 = 2 \sqrt 2 \sqrt 3 \sqrt 6 \)
    Use product rule of radicals to write \( \sqrt 2 \sqrt 3 \) as \( \sqrt 6 \) and substitute
    \( = 2 \sqrt 6 \sqrt 6 \)
    Use product rule of radicals to write \( \sqrt 6 \sqrt 6 = \sqrt {36} = 6 \) and simplify
    \( = 2 \times 6 = 12 \)



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