(a) Given that 110\(_x\) - 40\(_{five}\). find the value of x (b) Simplify \(\frac{15}{\sqrt{75}} + \(\sqrt{108}\) + \(\sqrt{432}\), leaving the answer in t...
Question 1 Report
(a) Given that 110\(_x\) - 40\(_{five}\). find the value of x
(b) Simplify \(\frac{15}{\sqrt{75}} + \(\sqrt{108}\) + \(\sqrt{432}\), leaving the answer in the form a\(\sqrt{b}\), where a and b are positive integers.
Answer Details
(a) Value of x. \(110_x = 40_{five}\).
Right side: \(40_{five}=4\times 5 + 0 = 20\).
Left side: \(110_x = 1\cdot x^{2} + 1\cdot x + 0 = x^{2}+x\).
\(x^{2}+x=20 \Rightarrow x^{2}+x-20=0 \Rightarrow (x+5)(x-4)=0\).
Since a base is positive, \(x=\mathbf{4}\).
(b) Simplify \(\dfrac{15}{\sqrt{75}}+\sqrt{108}+\sqrt{432}\).
\(\dfrac{15}{\sqrt{75}}=\dfrac{15}{5\sqrt{3}}=\dfrac{3}{\sqrt{3}}=\sqrt{3}\).
\(\sqrt{108}=\sqrt{36\times 3}=6\sqrt{3}\); \(\sqrt{432}=\sqrt{144\times 3}=12\sqrt{3}\).
Sum \(=\sqrt{3}+6\sqrt{3}+12\sqrt{3}=19\sqrt{3}\).
\(a=19,\ b=3\).
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