Question

Solve the cube root equation

\sqrt[3]{14+\sqrt{x} } +\sqrt[3]{14-\sqrt{x} } =4

Collected in the board: Algebraic equation

Steven Zheng posted 4 hours ago


Answer 1

The root of cubic equation z^3+pz+q=0 is


z=\sqrt[3]{-\dfrac{q}{2}+\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} } } +\sqrt[3]{-\dfrac{q}{2} -\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27} }}

The equation \sqrt[3]{14+\sqrt{x} } +\sqrt[3]{14-\sqrt{x} } =4 is identical to the root equation of the cubic equation. To find x is to determine the value of the expression \dfrac{q^2}{4}+\dfrac{p^3}{27}

Comparing the two equations gives,

-\dfrac{q}{2}=14 \to q=-28 Plug in the values of q and x to the cubic equation

4^3+4p-28=0 \to p=-9

Therefore,

x = \dfrac{q^2}{4}+\dfrac{p^3}{27} = \dfrac{28^2}{4}+\dfrac{(-9)^3}{27}=169

Steven Zheng posted 3 hours ago


Answer 2

Evaluating the cube root equation

\sqrt[3]{14+\sqrt{x} } +\sqrt[3]{14-\sqrt{x} } =4

Let

a = 14+\sqrt{x}
(1)
b=14 - \sqrt{x}
(2)

Then,

\sqrt[3]{a}+\sqrt[3]{b}=4
(3)

Addition (1) and(2) yields

a+b=28
(4)

Subtraction of (1) by (2) yields

a-b=2\sqrt{x}

x = \dfrac{(a-b)^2}{4}
(5)

Let

m = \sqrt[3]{a}
(6)
n=\sqrt[3]{b}
(7)

Then we obtain the equation system

m+n=4
(8)
m^3+n^3=28
(9)
(m+n)^3=m^3+3m^2n+3nm^2+n^3

=(m^3+n^3)+3mn(m+n)

Plug in (8) and(9) to above equation

4^3 = 28+3mn\cdotp 4

Solving the equation yields

mn = 3
(10)

Taking square of (8)

(m+n)^2 = 16

(m+n)^2-4mn = 16-4mn

(m-n)^2=16-4\times 3=4

m-n=\pm2

Since a >b, cancel negative value

m-n=2
(11)

Solving the equation system

m-n=2

m+n=4

we get

m=3, n=1

Then,

a=27, b = 1

Plug in to (5) to obtain

x=\dfrac{(a-b)^2}{4} =\dfrac{(27-1)^2}{4} =169

Steven Zheng posted 3 hours ago

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