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To include a square root sign in your answer use sqrt(). For example, to write $\\sqrt{3}$, type sqrt(3) into the answer box. If you are entering a number multiplied by the square root of some other number, for example $3\\sqrt{5}$, type 3*sqrt(5) into the answer box.

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Which of the following can be simplified further?

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Can be simplified further

", "

Cannot be simplified further

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$\\sqrt{\\var{p}}$

", "

$\\sqrt{\\simplify{{a}*{n}^2}}$

", "

$\\sqrt{\\var{a}}$

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Recall the  first rule of surds

\n

$\\sqrt{(ab)} = \\sqrt{a} \\times \\sqrt{b}$.

\n

\n

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"}, "advice": " #### a) \n Surds can be manipulated using the rule \n \$\\sqrt{(ab)} = \\sqrt{a} \\times \\sqrt{b}.\$ \n We are asked to state which of$\\sqrt{\\var{p}}$,$\\sqrt{\\simplify{{a}*{n}^2}}$, and$\\sqrt{\\var{a}}$can be simplified further. Commonly, surds can be simplified if the number inside of the square root has a square number as a factor. \n Here,$\\var{p}$is a prime number which means that its only divisors are$\\var{p}$and$1$. \n Therefore,$\\sqrt{\\var{p}}$cannot be simplified any further. \n Similarly,$\\var{a}$is also a prime number, so$\\sqrt{\\var{a}}$also cannot be simplified any further. \n On the other hand,$\\simplify{{a}*{n}^2}$is not a prime number and we can use the previous rule to simplify$\\sqrt{\\simplify{{a}*{n}^2}}as \n \\\begin{align} \\sqrt{\\simplify{{a}*{n}^2}} &= \\sqrt{\\simplify{{n}^2}} \\times \\sqrt{\\var{a}}\\\\ &= \\simplify{{n}*sqrt({a})}. \\end{align} \ \n #### b) \n Using the same rule of manipulation as in part a), we can simplify\\sqrt{\\simplify{{n}^2*{p}}}as \n \\\begin{align} \\sqrt{\\simplify{{n}^2*{p}}} &= \\sqrt{\\simplify{{n}^2}} \\times \\sqrt{\\var{p}}\\\\ &= \\simplify{{n}*sqrt({p})}. \\end{align} \ \n #### c) \n Here, we can use both of the rules for manipulating surds: \n \$\\sqrt{(ab)} = \\sqrt{a} \\times \\sqrt{b} \\text{.} \$ \n \$\\sqrt{\\frac{a}{b}} = \\frac{\\sqrt{a}}{\\sqrt{b}} \\text{.} \$ \n We can simplify\\displaystyle\\frac{ \\sqrt{\\simplify{{a}*{v}}} }{ \\sqrt{\\var{a}} }as follows. \n \\\begin{align} \\frac{\\sqrt{\\simplify{{a}*{v}}}}{\\sqrt{\\var{a}}} &= \\frac{\\sqrt{\\var{a}} \\times \\sqrt{\\var{v}}}{\\sqrt{\\var{a}}} \\\\[0.5em] &= \\frac{\\sqrt{\\var{a}}}{\\sqrt{\\var{a}}} \\times \\sqrt{\\var{v}} \\\\[0.5em] &= \\simplify{{sqrt(a)/sqrt(a)}} \\times \\sqrt{\\var{v}} \\\\[0.5em] &= \\sqrt{\\var{v}} \\text{.} \\end{align} \ \n Or, \n \\\begin{align} \\frac{\\sqrt{\\simplify{{a}*{v}}}}{\\sqrt{\\var{a}}} &= \\sqrt{\\frac{\\simplify{{a}*{v}}}{\\var{a}}} \\\\[0.5em] &= \\sqrt{\\var{v}} \\text{.} \\end{align} \ \n #### d) \n We can simplify the fraction as \n \\\begin{align} \\frac{\\sqrt{\\simplify{({b}{m})^2*{s}}}}{\\var{m}} &= \\frac{\\sqrt{\\simplify{({b*m})^2}} \\times \\sqrt{\\var{s}}}{\\var{m}} \\\\[0.5em] &= \\frac{\\simplify{{b*m}} \\times \\sqrt{\\var{s}}}{\\var{m}} \\\\[0.5em] &= \\simplify{{b}*sqrt({s})} \\text{.} \\end{align} \ \n #### e) \n \\\begin{align} \\simplify{{d}sqrt({a}) - {b}sqrt({v}^2{a})+{n}sqrt({b}^2*{a})} &= \\var{d}\\sqrt{\\var{a}} - \\var{b}(\\sqrt{\\simplify{{v}^2}} \\times \\sqrt{\\var{a}})+\\var{n}(\\sqrt{\\simplify{{b}^2}} \\times \\sqrt{\\var{a}}) \\\\ &= \\var{d}\\sqrt{\\var{a}} -\\var{b}(\\simplify{{v}*sqrt({a})})+\\var{n}(\\simplify{{b}*sqrt({a})}) \\\\ &= \\simplify{{d}sqrt({a})}-\\simplify{{b}*{v}sqrt({a})}+\\simplify{{n}*{b}sqrt({a})} \\\\ &= \\simplify{({d}-{b}*{v}+{n}*{b})sqrt({a})} \\text{.} \\end{align} \ \n #### f) \n We rationalise the denominator of fractions of the form\\displaystyle\\frac{1}{\\sqrt{a}}$, by multiplying the top and bottom by$\\sqrt{a}$. \n Therefore, to rationalise the denominator of the fraction$\\displaystyle\\frac{1}{\\sqrt{\\var{a}}}$, we multiply top and bottom by$\\sqrt{\\var{a}}. \n \\\begin{align} \\frac{1}{\\sqrt{\\var{a}}} &= \\frac{1}{\\sqrt{\\var{a}}} \\times \\frac{\\sqrt{\\var{a}}}{\\sqrt{\\var{a}}} \\\\[0.5em] &= \\frac{\\sqrt{\\var{a}}}{\\var{a}} \\text{.} \\end{align} \ \n #### g) \n We rationalise the denominator of fractions of the form\\displaystyle\\frac{1}{a+\\sqrt{b}}$by multiplying the top and bottom by$a-\\sqrt{b}$. \n Therefore, to rationalise the denominator of the fraction$\\displaystyle\\frac{1}{\\var{n}+\\sqrt{\\var{a}}}$, we multiply the top and bottom by$\\var{n} - \\sqrt{\\var{a}}. \n \\\begin{align} \\frac{1}{\\var{n}+\\sqrt{\\var{a}}} &= \\frac{1}{\\var{n}+\\sqrt{\\var{a}}} \\times \\frac{\\var{n}-\\sqrt{\\var{a}}}{\\var{n}-\\sqrt{\\var{a}}} \\\\[0.5em] &=\\frac{\\var{n}-\\sqrt{\\var{a}}}{(\\var{n}+\\sqrt{\\var{a}})(\\var{n}-\\sqrt{\\var{a}})} \\\\[0.5em] &=\\frac{\\var{n}-\\sqrt{\\var{a}}}{\\simplify{{n}^2}-\\var{a}} \\\\[0.5em] &=\\frac{\\var{n}-\\sqrt{\\var{a}}}{\\simplify{{n}^2-{a}}} \\text{.} \\end{align} \ \n #### h) \n We rationalise the denominator of fractions of the form\\displaystyle\\frac{1}{a-\\sqrt{b}}$by multiplying the top and bottom by$a+\\sqrt{b}$. \n Therefore, to rationalise the denominator of the fraction$\\displaystyle\\frac{\\var{t}}{\\var{d+p}-\\sqrt{\\var{p}}}$, we multiply the top and bottom by$\\var{d+p}+\\sqrt{\\var{p}}\$.

\n

\\\begin{align} \\frac{\\var{t}}{\\var{d+p}-\\sqrt{\\var{p}}} &= \\frac{\\var{t}}{\\var{d+p}-\\sqrt{\\var{p}}} \\times \\frac{\\var{d+p}+\\sqrt{\\var{p}}}{\\var{d+p}+\\sqrt{\\var{p}}} \\\\[0.5em] &=\\frac{\\var{t}(\\var{d+p}+\\sqrt{\\var{p}})}{(\\var{d+p}-\\sqrt{\\var{p}})(\\var{d+p}+\\sqrt{\\var{p}})} \\\\[0.5em] &=\\frac{\\var{t}(\\var{d+p}+\\sqrt{\\var{p}})}{\\simplify{{d+p}^2}-\\var{p}} \\\\[0.5em] &=\\frac{\\var{t}(\\var{d+p}+\\sqrt{\\var{p}})}{\\simplify{{d+p}^2-{p}}} \\\\[0.5em] &=\\simplify{{t}/{(d+p)^2-p}}(\\var{d+p}+\\sqrt{\\var{p}}) \\\\[0.5em] &= \\simplify[all,!noleadingMinus]{({t*(d+p)}+{t}*sqrt({p}))/({(d+p)^2-p})} \\text{.} \\end{align} \

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