Question

solve this question

$\sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6} } } }$
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Answer 1

Correct question: If x = $\sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6} } } }... \infty$. Find the value of x.

Answer:

$\large{\underline{\boxed{\sf x = 3}}}$

Step-by-step explanation:

Let x = $\sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6} } } }... \infty$

Thus,

$\sf x = \sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6} } } } ..... \infty \\ \\ \bf on \: squaring \: both \: sides : \\ \\ \implies \sf x {}^{2} = 6+x \\ \\ \implies \sf x {}^{2} - x - 6= 0$

We got a quadratic equation here, on comparing the given equation with ax² + bx + c, so

• a = 1
• b = - 1
• c = - 6

Discriminant = b² - 4ac

⇒ D = (-1)² - 4 * 1 * (-6)

⇒ D = 1 + 24

⇒ D = 25

$\tt x = \frac{ - b \pm \sqrt{D} }{2a} \\ \\ \implies \tt x = \frac{ - ( - 1) \pm \sqrt{25} }{2 \times 1} \\ \\ \implies \tt x = \frac{1 \pm 5}{2}$

Therefore,

$\implies \tt x = \frac{1+5}{2} = \frac{6}{2} = 3\\ \\ \implies \tt x = \frac{1-5}{2} = \frac{-4}{2} = - 2$

Thus, neglecting the negative value. We get x = 3.

Hence, the value of x is 3.