Question

Find The Value Of y


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Answer 1

Question :

$\bf{y = \sqrt[3]{6 + \sqrt[3]{6 + \sqrt[3]{6 + .\:.\:. \infin}}}}$

Solution :

By cubing the whole Equation, we get :

$:\implies \bf{y^{3} = \sqrt[3]{6 + \sqrt[3]{6 + \sqrt[3]{6 + .\:.\:. \infin}}}}$

$:\implies \bf{y^{3} = 6 + \sqrt[3]{6 + \sqrt[3]{6 + \sqrt[3]{6 + .\:.\:. \infin}}}}$

Let , $\bf{\sqrt[3]{6 + \sqrt[3]{6 + \sqrt[3]{6 + .\:.\:. \infin}}}}$ be y,

$:\implies \bf{y^{3} = 6 + y}$

$:\implies \bf{y^{3} - y - 6 = 0}$

Now , by using the trial method and substituting the value of y in the equation, we get :

$:\implies \bf{y^{3} - x - 6 = 0}$

$:\implies \bf{2^{3} - 2 - 6 = 0}$

$:\implies \bf{8 - 8 = 0}$

$:\implies \bf{\not{8} - \not{8} = 0}$

$:\implies \bf{0 = 0}$

Hence, the first root of the Equation is 2 or (y - 2 = 0).

Now by dividing the Equation (x³ - x - 6) by (x - 2) , we get :

⠀⠀⠀⠀⠀⠀⠀⠀⠀x - 2)x³ - x - 6(x² + 2x + 3

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ x³ - 2x²

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀----------

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀2x² - x

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀2x² - 4x

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀-----------

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀3x - 6

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀3x - 6

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀----------

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀0

Hence, the other root is (x² + 2x + 3).

Now , by using the formula for quadratic equation and substituting the values in it, we get :

Here,

• a = 1
• b = 2
• c = 3

Now , using the formula for quadratic equation and substituting the values in it, we get :

$:\implies \bf{y = \dfrac{-2 \pm \sqrt{2^{2} - 4 \times 1 \times 3}}{2 \times 1}}$

$:\implies \bf{y = \dfrac{-2 \pm \sqrt{4 - 12}}{2 \times 1}}$

$:\implies \bf{y = \dfrac{-2 \pm \sqrt{(-8)}}{2}}$

$\boxed{\therefore \bf{y = \dfrac{-2 \pm \sqrt{(-8)}}{2}}}$

Since , the roots of the other Equation i.e, x² + 2x + 3 is not forming a perfect root , the original value of y will be 2.

Answer 2

Question -

Find the value of y in the following question

$\sf y = \sqrt[3]{6+\sqrt[3]{6 +\sqrt[3]{6+...\infty}}}$

Answer -

$\sf y = \sqrt[3]{6+\sqrt[3]{6 +\sqrt[3]{6+...\infty}}}$

Let , $\sf y = \sqrt[3]{6+\sqrt[3]{6 +\sqrt[3]{6+...\infty}}}$

Substituting the value of y in the given Question we get ,

$\implies\:y\:=\:\sqrt{3}[6\:+\:(\:y\:)]$

On cubing both the sides ,We have the following statement

,$\implies\:y^3\:=\:6\:+\:y$

We know that a equation is always equal to zero . So thus we have ,

$\implies\:y^3\:-\:y\:-\:6\:=\:0$

On factorising this equation we have ,

$\implies\:y^3\:-\:4y\:+\:3y\:-\:6\:=\:0$

Taking out y and 3 as common we get ,

$\implies\:y\:\times\:(\:y^2\:-\:4\:)\:+\:3\:\times\:(\:y\:-\:2\:)$

$\implies\:y\:(\:y\:-\:2\:)\:\times\:(\:y\:+\:2\:)\:+\:3\:\times\:(\:y\:-\:2\:)$

$\implies\:(\:y\:-\:2\:)(\:y^2\:+\:2y\:+\:3)\:=\:0$

Given is that

$\implies\:(\:y\:-\:2\:)\:=\:0$

$(\:y^2\:+\:2y\:+\:3)\:=\:0$

Solving the equation for the value of y ,

$\implies\:(\:y\:-\:2\:)\:=\:0$

$\implies\:y\:=\:2\:+\:0$

$\implies\:y\:=\:2$

Another method

Now let's find the value of y by trial method ,

$\implies\:y^3\:-\:y\:-\:6\:=\:0$

Let y be 1 , so we have

$\implies\:1^3\:-\:1\:-\:6\:=\:0$

$\implies\:1\:-\:1\:-\:6\:=\:0$$\implies\:-\:6\:\cancel{=}\:0$

LHS is not equal to RHS .

Let y be 2 so now we have ,

$\implies\:2^3\:-\:2\:-\:6\:=\:0$

$\implies\:8\:-\:2\:-\:6\:=\:0$

$\implies\:8\:-\:8\:=\:0$

$\implies\:0\:=\:0$

Hence , proved that the value of y is 2 in the Question $\sf y = \sqrt[3]{6+\sqrt[3]{6 +\sqrt[3]{6+...\infty}}}$

Why can't be 3 or - 2 be the the solutions of this equation ?

Here's why . Let the value of y be -2

$\implies\:-2^3\:-\:(-2)\:-\:6\:=\:0$

$\implies\:-8\:+\:2\:-\:6\:=\:0$

$\implies\:-8\:-\:4\:=\:0$

$\implies\:-12\:\cancel{=}\:0$

Let the value of y be 3 .

$\implies\:3^3\:-\:3\:-\:6\:=\:0$

$\implies\:27\:-\:3\:-\:6\:=\:0$

$\implies\:18\:\cancel{=}\:0$

Hence these values can't be the solutions