Solutions to Simplify Radicals Questions

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

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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