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Question Number 154444 by Rasheed.Sindhi last updated on 18/Sep/21

Commented by Rasheed.Sindhi last updated on 18/Sep/21

$You can't use 'macro parameter character #' in math mode$
	Answered by Rasheed.Sindhi last updated on 18/Sep/21

	$\sqrt{5 x + 1} + \sqrt{x - 2} = \sqrt{8 x + 1}; S olve .$ $\underset{a}{\underset{⏟}{\sqrt{\frac{8 x + 1}{x - 2}}}} - \underset{b}{\underset{⏟}{\sqrt{\frac{5 x + 1}{x - 2}}}} = 1$ $a - b = 1 . . . . . . . . . . A$ $a^{2} - b^{2} = {(\sqrt{\frac{8 x + 1}{x - 2}})}^{2} - {(\sqrt{\frac{5 x + 1}{x - 2}})}^{2}$ $a^{2} - b^{2} = \frac{3 x}{x - 2} . . . . . . . . . B$ $B / A : a + b = \frac{3 x}{x - 2} . . . . . . C$ $A + C : 2 a = \frac{3 x}{x - 2} + 1 = \frac{4 x - 2}{x - 2}$ $a = \frac{2 x - 1}{x - 2}$ $\frac{8 x + 1}{x - 2} = {(\frac{2 x - 1}{x - 2})}^{2}$ $8 x + 1 = \frac{4 x^{2} - 4 x + 1}{x - 2}$ $8 x^{2} + x - 16 x - 2 = 4 x^{2} - 4 x + 1$ $4 x^{2} - 11 x - 3 = 0$ $4 x^{2} - 12 x + x - 3 = 0$ $4 x (x - 3) + 1 (x - 3) = 0$ $(x - 3) (4 x + 1) = 0$ $x = 3 (✓) ∣ x = - \frac{1}{4} (f a l s e)$
	Answered by cherokeesay last updated on 18/Sep/21
	$5x+1+x−2+2(√((5x+1)(x−2))) = 8x+1 (x≥2)⇔x∈[2 , +∞[ ⇒ (√((5x+1)(x−2))) = x+1 ⇔ (5x+1)(x−2)=(x+1)^2 ⇒ 4x^2 −11x−3=0 ⇔(4x+1)(x−3)=0 x = 3 or x = −(1/4) ∉[2 , +∞[ S={3}$
	$5 x + 1 + x - 2 + 2 \sqrt{(5 x + 1) (x - 2)} = 8 x + 1$ $(x ⩾ 2) \Leftrightarrow x \in [2, + \infty [$ $\Rightarrow \sqrt{(5 x + 1) (x - 2)} = x + 1 \Leftrightarrow$ $(5 x + 1) (x - 2) = {(x + 1)}^{2} \Rightarrow$ $4 x^{2} - 11 x - 3 = 0 \Leftrightarrow (4 x + 1) (x - 3) = 0$ $x = 3 o r x = - \frac{1}{4} \notin [2, + \infty [$ $S = {3}$
	Commented by Rasheed.Sindhi last updated on 19/Sep/21

	$T han k s ser!$
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