How sharp are your eyes part 1

Algebra Level 4

Find the product of the roots of all real $x$ satisfying the equation $\sqrt[5]{(7x-3)^3}+8\sqrt[5]{(3-7x)^{- 3}}=7$

\frac{23}{5}

\frac{10}{7}

\frac{17}{3}

\frac{12}{7}

1 solution

P C
Dec 30, 2015

Condition of x: $x\neq\frac{3}{7}$ Let $\sqrt[5]{(7x-3)^3}=a$ We get $a-\frac{8}{a}=7$ $\Leftrightarrow a^2-8=7a$ $\Leftrightarrow a^2-7a-8=0$ Solve the above equation and you get $a=8\Leftrightarrow x=5$ $a=-1\Leftrightarrow x= \frac{2}{7}$ $\Rightarrow 5.\frac{2}{7} = \frac{10}{7}$

You should mention that x is real, otherwise other solutions can exist. For example:If $\sqrt[5]{(7x - 3)^{3}} = -1 \implies (7x - 3) = \sqrt[3]{-1} \implies 7x - 3 = -1, -\omega, -\omega^{2} \implies, x = \frac{2}{7}, \frac{3 - \omega}{7}, \frac{3 - \omega^{2}}{7}$ . Similarly, when $\sqrt[5]{(7x - 3)^{3}} = 8$ .

Krutarth Patel - 5 years, 1 month ago

How sharp are your eyes part 1

1 solution

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