Question

CA foundation question. Solve with required steps.

Thanks in advance.​

Answer 1

Topic - Infinite Tatration

Explanation:

We need to solve the given expression.

$\displaystyle \rm 4 + \dfrac{1}{4 + \dfrac{1}{4 + \dfrac{1}{4 + ... \infty} } }$

Let's say the given expression be some variable say (y)

${ \displaystyle \implies \rm y = 4 + \dfrac{1}{4 + \dfrac{1}{4 + \dfrac{1}{4 + ... \infty} } } }$

Can be re-written as,

${ \displaystyle \implies \rm y = 4 + \dfrac{1}{y}}$

${ \displaystyle \implies \rm y = \dfrac{4y + 1}{y}}$

${ \displaystyle \implies \rm {y}^{2} = 4y + 1}$

${ \displaystyle \implies \rm {y}^{2} - 4y - 1 = 0 \: \: \: ... [Eq. 1]}$

Now, we will use quadratic formula to solve for irrational roots of this equation.

According to quadratic formula:

For any equation of the form $ax^2 + bx+c = 0$, it's roots are given by the below formula.

$\boxed{x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a} }$

By using this formula, the roots of [Eq. 1] are given by:

${ \displaystyle \cdot\cdot\cdot\longrightarrow \rm y = \frac{ - ( - 4) \pm \sqrt{ {( - 4)}^{2} -4(1)( - 1) } }{2} }$

${ \displaystyle\cdot\cdot\cdot\longrightarrow \rm y = \frac{ 4 \pm \sqrt{16 + 4} }{2} }$

${ \displaystyle \cdot\cdot\cdot\longrightarrow\rm y = \frac{ 4 \pm \sqrt{20} }{2} }$

${ \displaystyle\cdot\cdot\cdot\longrightarrow \rm y = \frac{ 4 \pm \sqrt{2 \cdot 2 \cdot 5} }{2} }$

${ \displaystyle\cdot\cdot\cdot\longrightarrow \rm y = \frac{ 4 \pm 2\sqrt{5} }{2} }$

${ \displaystyle \cdot\cdot\cdot\longrightarrow\rm y = \frac{2( 2 \pm \sqrt{5}) }{2} }$

${ \displaystyle\cdot\cdot\cdot\longrightarrow \rm y = 2 \pm \sqrt{5}}$

Therefore, option [C] is correct answer.

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Answer 2

Step-by-step explanation:

We know the formula,

$\\ \tt\[ \frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \]$

Here the equation,

$\\ \\ \[ \begin{array}{l} \tt 2 x^{2}-4 x-3=0 \\ \\ \\ \tt a=2, b=-4, c=-3 \end{array} \]$

Substituting the values, we get

$\\ \\ \[ \begin{array}{l} \tt =\dfrac{4 \pm \sqrt{(-4)^{2}-4 \times 2 \times-3}}{2 \times 2} \\ \\ \\ \tt=\dfrac{4 \pm \sqrt{16+24}}{4} \\ \\ \\ \tt =\cfrac{4 \pm \sqrt{40}}{4} \end{array} \]$

Take 2 as common, we get

$\\ \\ \tt\[ =\frac{2 \pm \sqrt{10}}{2} \]$

$\\ \\ \tt=$

$\\ \\ \color{maroon} \text{\( \therefore \) solution of the equation \( \tt2 x^{2}-4 x-3 \) is \( \tt \dfrac{2 \pm \sqrt{10}}{2} \).}$