Solve-for-x-x-2-c-c-x-

Question Number 193130 by Frix last updated on 04/Jun/23

Solve for x

x^{2} - c = \sqrt{c - x}

Commented by aba last updated on 04/Jun/23

t = c - x ⩾ 0

x^{2} - (t - x) = \sqrt{t} \Rightarrow x^{2} - x - (t + \sqrt{t}) = 0

Δ = {(2 \sqrt{t} + 1)}^{2}

x_{1} = \frac{1 - (2 \sqrt{t} + 1)}{2} = - \sqrt{t} \land x_{2} = \frac{1 + (2 \sqrt{t} + 1)}{2} = 1 + \sqrt{t}

⧫ x = - \sqrt{c - x} \Rightarrow x^{2} + x - c = 0

Δ = 1 + 4 c > 0

x = \pm \frac{1}{2} (\sqrt{4 c + 1} - 1) ✓

⧫ x = 1 + \sqrt{c - x} \Rightarrow x - 1 = \sqrt{c - x} \Rightarrow x^{2} - x + (1 - c) = 0

Δ = 1 + 4 (c - 1) = 4 c - 3 > 0

x_{1} = \frac{1 - \sqrt{4 c - 3}}{2} \land x = \frac{1 + \sqrt{4 c - 3}}{2} ✓

S = {\frac{1 + \sqrt{4 c - 3}}{2}; \pm \frac{1}{2} (\sqrt{4 c + 1} - 1)}

Answered by Mastermind last updated on 04/Jun/23

Solution

take square of both sides, we have

\Rightarrow {(x^{2} - c)}^{2} = c - x

\Rightarrow x^{4} - {2 x}^{2} c + c^{2} = c - x

\Rightarrow c^{2} - c - {2 x}^{2} c + x^{4} - x = 0

\Rightarrow c^{2} - (1 + {2 x}^{2}) + x^{4} - x = 0 \cdot \cdot \cdot \cdot \cdot *

\Rightarrow solve equation * by using Almighty

formular (Quadratic formula)

c = \frac{- b \pm \sqrt{b^{2} - 4 ac}}{2 a}

where b = - (1 + {2 x}^{2}), a = 1 and c = x^{4} - 4

therefore,

c = \frac{(1 + {2 x}^{2}) \pm \sqrt{(1 + {4 x}^{2} + {4 x}^{4}) - 4 (x^{4} - x)}}{2 \times 1}

c = \frac{(1 + {2 x}^{2})}{2} \pm \frac{\sqrt{1 + {5 x}^{2}}}{2}

Answered by aba last updated on 04/Jun/23

x = \pm \frac{1}{2} (\sqrt{4 c + 1} - 1) \lor x = \frac{1}{2} (\sqrt{4 x - 3} + 1)

Answered by Frix last updated on 04/Jun/23

Finding one solution using f (x) = f^{- 1} (x)

y = \sqrt{c - x} \Leftrightarrow x = c - x^{2} = - (x^{2} - c)

\Rightarrow

- x = \sqrt{c - x} \Leftrightarrow x^{2} = c - x \land x < 0 \Rightarrow

x_{1} = - \frac{1 + \sqrt{4 c + 1}}{2} [x_{2} = - \frac{1 - \sqrt{4 c + 1}}{2} ⩾ 0]

x^{2} - c = \sqrt{c - x}

parabola = positive half parabola \Rightarrow 2 solutions

Solving for c is easy

{(x^{2} - c)}^{2} = c - x

c^{2} - (2 x^{2} + 1) c + x (x^{3} + 1) = 0

c = \frac{2 x^{2} + 1}{2} \pm \frac{2 x - 1}{2}

c = x^{2} - x + 1 \lor c = x^{2} + x

Now solve these for x

x = \frac{1 \pm \sqrt{4 c - 3}}{2} \lor x = - \frac{1 \pm \sqrt{4 c + 1}}{2}

But 2 solutions are false

x_{1} = - \frac{1 + \sqrt{4 c + 1}}{2} [from above]

x_{2} = \frac{1 + \sqrt{4 c - 3}}{2} [because it must be > 0 due

to symmetry of x^{2} - c]

Commented by aba last updated on 04/Jun/23

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