Discussion

Question:

The product of the roots of the equation $\sqrt[3]{11 + 2 x}$ + $\sqrt[3]{11 - 2 x}$ = 2 is approximately equal to:

Explanation:

Cubing the given equation, we have:

(a + b)³ = a³ + b³ + 3 ∙ a ∙ b(a + b)

11 + 2x + 11 - 2x + 3 ∙ ${(11 + 2 x)}^{\frac{1}{3}}$ ${(11 - 2 x)}^{\frac{1}{3}}$ (a + b)

Given a + b = 2.

Hence:

22 + 3 ∙ ${(11 + 2 x)}^{\frac{1}{3}}$ ${(11 - 2 x)}^{\frac{1}{3}}$ (2)

= 22 + 6 × $({(121 - 4 x^{2})}^{\frac{1}{3}})$ = 8

Cubing this we get: 121 - 4x² = $- \frac{343}{27}$

4x² = 121 + $\frac{343}{27}$ = $\frac{3610}{27}$

x² = 33.42

One root would be + $\sqrt{33.42}$ and the other - $\sqrt{33.42}$ .

The product of the roots would be approximately -33.

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