The reunion of broken parts

Algebra Level 2

$\large x \cdot \sqrt{x^2} \cdot \sqrt[3]{x^3} \cdot \sqrt[4]{x^4} = 81$

Which of the following options is true about $x^\frac 32$ , assume that $x>0$ .

x^{\frac{3}{2}} \in \mathbb{Z}

x^{\frac{3}{2}} \in \mathbb{N}

x^{\frac{3}{2}}\in \mathbb{R}

and

x^{\frac{3}{2}} \in \mathbb{Q}

x^{\frac{3}{2}} \in \mathbb{R}

and

x^{\frac{3}{2}} \in \mathbb{C}

x^{\frac{3}{2}} \in \mathbb{Q}

1 solution

Munem Shahriar
Jan 14, 2018

$x \times \sqrt{x^2} \times \sqrt[3]{x^3} \times \sqrt[4]{x^4} = 81$

$x \times x \times x \times x = 81$

$x^4 = 81$

$x = \sqrt[4]{81}$

$\implies x = 3$

Hence $x^{3/2} = 3^{3/2} = 3\sqrt3$ which is a real number and also a complex number.

Note: $\sqrt3$ is an irrational number.

Actually $x=-3$ is also a solution.

$x \cdot \sqrt{x^2} \cdot \sqrt[3]{x^3} \cdot \sqrt[4]{x^4} =-3 \cdot \sqrt{(-3)^2} \cdot \sqrt[3]{(-3)^3} \cdot \sqrt[4]{(-3)^4} = -3\cdot|-3|\cdot(-3)\cdot|-3|=81$

Marta Reece - 3 years, 4 months ago

Problem was updated to assume x>0

Jerry McKenzie - 3 years, 4 months ago

The reunion of broken parts

1 solution

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