How do you solve $5 \sqrt[3]{4 x - 1} + 2 = 17$ $5root3(4x-1)+2=17$ ?

Question

1948 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-01T06:43:32

$x = 7$ $x=7$

$5 \times \sqrt[3]{4 x - 1} + 2 = 17$ $5xxroot(3)(4x-1)+2=17$

Here, we try to keep the cube root term on one side.

Subtract $2$ $2$ from both sides:

$5 \times \sqrt[3]{4 x - 1} + 2 - 2 = 17 - 2$ $5xxroot(3)(4x-1)+2color(red)(-2)=17color(red)(-2)$

$5 \times \sqrt[3]{4 x - 1} = 15$ $5xxroot(3)(4x-1)=15$

Next, divide both sides by $5$ $5$ :

$\frac{5 \times \sqrt[3]{4 x - 1}}{5} = \frac{15}{5}$ $[5xxroot(3)(4x-1)]/color(red)(5)=15/color(red)(5)$

$\sqrt[3]{4 x - 1} = 3$ $root(3)(4x-1)=3$

Now, we remove the cube root.
In order to do so, find the cube of both sides:

${(\sqrt[3]{4 x - 1})}^{3} = 3^{3}$ $(root(3)(4x-1))^color(red)(3)=3^color(red)(3)$

$4 x - 1 = 3 \times 3 \times 3$ $4x-1=3xx3xx3$

$4 x - 1 = 27$ $4x-1=27$

$4 x - 1 + 1 = 27 + 1$ $4x-1color(red)(+1)=27color(red)(+1)$

$4 x = 28$ $4x=28$

$\frac{4 x}{4} = \frac{28}{4}$ $(4x)/color(red)(4)=28/color(red)(4)$

$x = 7$ $x=7$

How do you solve 5root3(4x-1)+2=1753√4x−1+2=175root3(4x-1)+2=17?