Question

If x is a positive number such that the ratio 3x to
$\sqrt[3]{x}$

is equal to 27:1 then find then find the value of

${x}^{3} \ratio27 \sqrt[3]{x {}^{2} }$
please help
​

Answer 1

Step-by-step explanation:

given :

If x is a positive number such that the ratio 3x to

to find :

is equal to 27:1 then find then find the value of

solution :

• → The given equation 3x² + 2kx + 27 = 0

• Here, a = 3, b = 2k, c = 27

• It is given that roots are real and equal.

• ... b² - 4ac0

• ⇒ (2k)² - 4(3) (27) = 0

• ⇒ 4k².

• 324 0

• ⇒ 4k²

• ⇒ k² = 81

• = 324

• .. k = ±9

Answer 2

Given data:

The ratio $3x:\sqrt[3]{x}=27:1$

To find:

The ratio $x^{3}:27\sqrt[3]{x^{2}}$

Step-by-step explanation:

Let us find the value of $x$ first and later in the solution, we will use it.

Given, $3x:\sqrt[3]{x}=27:1$

$\Rightarrow \dfrac{3x}{\sqrt[3]{x}}=\dfrac{27}{1}$

$\Rightarrow \dfrac{x}{\sqrt[3]{x}}=\dfrac{9}{1}$

Taking the power $3$ in both sides, we get

$\quad \dfrac{x^{3}}{x}=\dfrac{729}{1}$

$\Rightarrow x^{2}=729$

$\Rightarrow \bold{x=27}$

Now, $x^{3}:27\sqrt[3]{x^{2}}$

$=27^{3}:27\sqrt[3]{27^{2}}$

$=27^{3}:27\sqrt[3]{9^{3}}\quad[\because 27^{2}=9^{3}]$

$=27^{3}:27\times 9$

$=27^{2}:9$

$=729:9$

$=81:1$

Final answer: $x^{3}:27\sqrt[3]{x^{2}}=81:1$

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