Question-194363

Question Number 194363 by sonukgindia last updated on 05/Jul/23

Commented by Frix last updated on 05/Jul/23

{There}^{'} s no solution .

\sqrt{z_{1} + \sqrt{z_{2}}} + \sqrt{z_{1} - \sqrt{z_{2}}} = r

Solving by 2 times [squaring (introduces

f alse solutions) and transforming] leads to :

z_{1} = \frac{r^{4} + 4 z_{2} + r^{4}}{4 r^{2}}

z_{1} = a + b i \land z_{2} = c + d i with a, b, c, d \in R

a + b i = \frac{4 c + r^{4}}{4 r^{2}} + \frac{d}{r^{2}} i

{\begin{cases} a = \frac{4 c + r^{4}}{4 r^{2}} \\ b = \frac{d}{r^{2}} \end{cases}

We have r = 2 \land c = a + 17

\Rightarrow a = 7 \land d = 4 b \Rightarrow

f (b) = \sqrt{7 + b i + 2 \sqrt{6 + b i}} + \sqrt{7 + b i - 2 \sqrt{6 + b i}}

The real part of f (b) is ⩾ 2 \sqrt{6}

The imaginary part of f (b) has only one

zero at b = 0

\Rightarrow f (0) = 2 \sqrt{6} is the only real point of f (b)

Answered by MacX last updated on 05/Jul/23

\sqrt{z + \sqrt{z + 17}} + \sqrt{z - \sqrt{z + 17}} = 2

\Rightarrow z + \sqrt{z + 17} + z - \sqrt{z + 17} + 2 \sqrt{z^{2} - (z + 17)} = 4

\Rightarrow 2 z + 2 \sqrt{z^{2} - z - 17} = 4

\Rightarrow 2 \sqrt{z^{2} - z - 17} = 4 - 2 z = 2 (2 - z)

\Rightarrow z^{2} - z - 17 = z^{2} - 4 z + 4

\Rightarrow - z - 17 = - 4 z + 4

\Rightarrow 3 z = 21

\Rightarrow z = 7

S_{C} = {7}

Commented by Frix last updated on 05/Jul/23

\sqrt{7 + \sqrt{7 + 17}} + \sqrt{7 - \sqrt{7 + 17}} = \sqrt{7 + \sqrt{24}} + \sqrt{7 - \sqrt{24}} =

= (1 + \sqrt{6}) + (- 1 + \sqrt{6}) = 2 \sqrt{6} \neq 2

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