(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.
Explanation
(a)110\(_x\) - 40\(_{5}\)
1 x \(x^2}\) + 1 x \(x^1\) + 0 x \(x^o\)
x = 4 x 5\(^1\) + 0 x 5\(^o\)
x\(^2\) + x = 4 x 5 + 0
x\(^2\) + x - 20 = 0
(x - 4)(x + 5) = 0
x - 4 = 0 or x + 5 = 0
x = 4 or x = -5
x = 4
(b) Simplify \(\frac{15}{\sqrt{75}}\) + \(\sqrt{108}\) + \(\sqrt{432}\)
= \(\frac{15}{\sqrt{25 \times 3}}\) + \(\sqrt{36 \times 3}\)
= \(\frac{15}{5 \sqrt{3}\) + 6\(\sqrt{3}\) + 12\(\sqrt{3}\)
= \(\frac{15 \sqrt{3}}{15 \times \sqrt{3} \times \sqrt{3}}\) + 18\(\sqrt{3}\)
= \(\frac{15\sqrt{3}}{15}\) + 18\(\sqrt{3}\)
\(\sqrt{3}\) + 18\(\sqrt{3}\) = 19\(\sqrt{3}\)