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Solutions 1. \( x^{-\frac{3}{2}}=8 \) \[ \begin{aligned} \left(x^{-\frac{3}{2}}\right)^{-\frac{2}{3}} & =\left(2^{3}\right)^{-\frac{2}{3}} \quad \checkmark\left(\text { raise both sides to the power of }-\frac{2}{3}\right) \\ x^{+1} & =2^{-2} \quad(2)^{3 \times-\frac{2}{3}} \\ x & =\frac{1}{4} \end{aligned} \] 2. \( \sqrt[5]{x^{4}}=256 \quad \) (change radical to fractional exponent form) \[ \begin{aligned} x^{\frac{4}{5}} & = \pm(2)^{8} \quad \quad \text { (raise both sides of equation to the power of } \frac{5}{4} \text { ) } \\ \left(x^{\frac{4}{5}}\right)^{\frac{5}{4}} & = \pm\left(2^{8}\right)^{\frac{5}{4}} \checkmark \\ \therefore x & = \pm(2)^{10} \checkmark= \pm 1024 \end{aligned} \] [7] Solutions 1. \( \sqrt{3 x+4}-5=0 \) \( \sqrt{3 x+4}=5 \) (isolate the radical) \( (\sqrt{3 x+4})^{2}=5^{2} \quad \) (square both sides of the equation) \( 3 x+4=25 \) \( 3 x=21 \) \[ x=7 \] 2. \( \sqrt{3 x-5}-x=5 \) \[ \begin{aligned} \sqrt{3 x-5} & =x-5 \\ (\sqrt{3 x-5})^{2} \Omega & =(x-5)^{2} \\ 3 x-5 & =x^{2}-10 x+25 \\ 0 & =x^{2}-13 x+30 \\ 0 & =(x-10)(x-3) \\ x & =10 \text { or } x=3 \end{aligned} \] Check your answer: If \( x=10 \) LHS: \( \sqrt{3(10)-5}-10 \) Check: LHS: \( \sqrt{3(7)+4}-5 \) \[ =\sqrt{21+4}-5 \] \[ =\sqrt{25}-5 \] \[ =0 \] \[ =\text { RHS } \] \( \therefore x=7 \) is a solution (always isolate the radical first) (square both sides) Remember: \( (x-5)^{2} \neq x^{2}+25 \) (quadratic equation, set \( =0 \) ) (factorise the trinomial and make each factor \( =0 \) )

Ask by Davison Reid. in South Africa
Mar 15,2025

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Solutions: 1. \( x = \frac{1}{4} \) 2. \( x = 1024 \) 3. \( x = 7 \) 4. No solution

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Did you know that the concept of exponents can be traced back to ancient civilizations? The Egyptians used a form of exponentiation in their calculations about fractions, while the Greeks, like Euclid, laid foundational principles in mathematics that eventually led to the formal definition of exponents and roots used in algebra today. The notation itself was later popularized in the 1600s, making it much easier for mathematicians to express these powerful concepts without lengthy explanations. When solving equations with radicals, it's crucial to keep an eye out for extraneous solutions that may arise during the squaring process. A common mistake is forgetting to check answers in the original equation, which can lead you to conclude incorrect solutions that satisfy the modified equation but not the initial one. Always isolate the radical and check each potential solution by substituting it back into the original equation to validate it. Happy solving!

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