Question-187835

Question Number 187835 by pascal889 last updated on 22/Feb/23

Commented by Frix last updated on 22/Feb/23

\sqrt{x - a} + \sqrt{x - b} = \sqrt{c}

x = \frac{{(a - b)}^{2}}{4 c} + \frac{a + b}{2} + \frac{c}{4}

Answered by HeferH last updated on 22/Feb/23

x = 2

Answered by Rasheed.Sindhi last updated on 22/Feb/23

\sqrt{x - 1} + \sqrt{x - 2} = 1

(\sqrt{x - 1} + \sqrt{x - 2}) (\sqrt{x - 1} - \sqrt{x - 2})

= 1 (\sqrt{x - 1} - \sqrt{x - 2})

\sqrt{x - 1} - \sqrt{x - 2} = {(\sqrt{x - 1})}^{2} - {(\sqrt{x - 2})}^{2}

= x - 1 - x + 2 = 1

{\begin{cases} \sqrt{x - 1} + \sqrt{x - 2} = 1 \\ \sqrt{x - 1} - \sqrt{x - 2} = 1 \end{cases}

\Rightarrow 2 \sqrt{x - 1} = 2 \Rightarrow x - 1 = 1 \Rightarrow x = 2

Answered by Rasheed.Sindhi last updated on 22/Feb/23

\sqrt{x - 1} = a, \sqrt{x + 2} = b

a + b = 1 \dots . . A

a^{2} - b^{2} = {(\sqrt{x - 1})}^{2} - {(\sqrt{x - 2})}^{2}

= x - 1 - x + 2 = 1

\frac{a^{2} - b^{2}}{a + b} = \frac{1}{1}

a - b = 1 \dots . . B

A + B : 2 a = 2 \Rightarrow a = 1

\sqrt{x - 1} = 1 \Rightarrow x - 1 = 1 \Rightarrow x = 2

Answered by manxsol last updated on 22/Feb/23

\sqrt{x - a} + \sqrt{x - b} = \sqrt{c} I

Extra \left or missing \right

\sqrt{x - a} - \sqrt{x - b} = \frac{b - a}{\sqrt{c}} I I

+ I + I I

2 \sqrt{x - a} = \sqrt{c} + \frac{b - a}{\sqrt{c}}

2 \sqrt{c} \sqrt{x - a} = (c + b - a)

4 c (x - a) = {(c + b - a)}^{2}

c o n d i t i o n

x ⩾ a

x ⩾ b

c + b ⩾ a

e x e r c i s e

a = 1

b = 2

c = 1

4 c (x - a) = {(c + b - a)}^{2}

4 (1) (x - 1) = {(2 + 1 - 1)}^{2}

4 (x - 1) = 4

x - 1 = 1

x = 2

Answered by SEKRET last updated on 23/Feb/23

\sqrt{x - 1} = a \sqrt{x - 2} = b

{\begin{cases} a + b = 1 \\ a - b = 1 \end{cases} a = 1 b = 0 x = 2

Commented by manxsol last updated on 23/Feb/23

e x c e l l e n t

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