Solve the equation: (√(5x^2 +14x+9))−(√(x^2 −x−20))=5(√(x+1))


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Question Number 197753 by tri26112004 last updated on 27/Sep/23

$S o l v e t h e e q u a t i o n :$ $\sqrt{5 x^{2} + 14 x + 9} - \sqrt{x^{2} - x - 20} = 5 \sqrt{x + 1}$
	Answered by Frix last updated on 27/Sep/23

	$Squaring, transforming, solving, testing$ $\Rightarrow$ $x_{1} = 8$ $x_{2} = \frac{5 + \sqrt{61}}{2}$
	Commented by tri26112004 last updated on 27/Sep/23

	$c a n y o u s h o w y o u r s o l u t i o n$
	Commented by Frix last updated on 27/Sep/23

	$\sqrt{a} - \sqrt{b} = \sqrt{c}$ $\sqrt{a} = \sqrt{b} + \sqrt{c}$ $a = b + 2 \sqrt{b c} + c$ $a - (b + c) = 2 \sqrt{b c}$ ${(a - (b + c))}^{2} = 4 b c$ $a^{2} + b^{2} + c^{2} - 2 (a b + a c + b c) = 0$ $Now insert$ $a = 5 x^{2} + 14 x + 9$ $b = x^{2} - x - 20$ $c = 25 (x + 1)$ $x^{4} - \frac{45 x^{3}}{4} + \frac{33 x^{2}}{4} + \frac{505 x}{4} + 126 = 0$ $(x - 8) (x + \frac{7}{4}) (x^{2} - 5 x - 9) = 0$ $Test the solutions, some might be false due$ $to squaring$
	Commented by tri26112004 last updated on 28/Sep/23

	$D o y o u h a v e a n o t h e r s o l u t i o n ¿$
	Commented by tri26112004 last updated on 28/Sep/23

	$S u c h a s U s i n g I n e q u a l i t y$
	Answered by Rasheed.Sindhi last updated on 28/Sep/23

	$\sqrt{5 x^{2} + 14 x + 9} - \sqrt{x^{2} - x - 20} = 5 \sqrt{x + 1}$ $x^{2} - x - 20 ⩾ 0 \land 5 x^{2} + 14 x + 9 ⩾ 0 \land x + 1 ⩾ 0$ $(x - 5) (x + 4) ⩾ 0$ $x - 5 ⩾ 0 \land x + 4 ⩾ 0$ $x ⩾ 5 \land x ⩾ - 4 \Rightarrow x ⩾ 5 . . . . . . . (i)$ $5 x^{2} + 14 x + 9 ⩾ 0$ $(x + 1) (5 x + 9) ⩾ 0$ $x ⩾ - 1 \land x ⩾ - \frac{9}{5} \Rightarrow x ⩾ - 1 . . . . (i i)$ $x + 1 ⩾ 0 \Rightarrow x ⩾ - 1 . . . . . . . . . . . . (i i i)$ $(i), (i i) & (i i i) :$ $x ⩾ 5 \land x ⩾ - 1$ $\Rightarrow x ⩾ 5$ $\sqrt{5 x^{2} + 14 x + 9} > \sqrt{x^{2} - x - 20}$ $5 x^{2} + 14 x + 9 > x^{2} - x - 20$ $4 x^{2} + 15 x + 29 > 0$
	Commented by tri26112004 last updated on 28/Sep/23

	$C a n y o u u s e t h e i n e q u a l i t y ¿$
	Commented by Rasheed.Sindhi last updated on 28/Sep/23

	$I^{'} v e t r i e d t h a t b u t f a i l e d t o g e t e x a c t a n s w e r .$
	Commented by tri26112004 last updated on 28/Sep/23

	$B u t i t h a s s o m e b e a u t y r o o t$ $x = 8 . . .$
	Commented by Frix last updated on 28/Sep/23

	$The solution method {doesn}^{'} t care whether$ $you like the solution or not .$
	Commented by tri26112004 last updated on 28/Sep/23

	$o k a y$
	Answered by ajfour last updated on 28/Sep/23
	$5x^2 +14x+9=5(x^2 −x−20) +19(x+1)+90 (√a)−(√b)=5(√c) a=5b+19c+90 b=c^2 −3c−18 ⇒ a−b=4b+19c+90 Also a=5(x+1)^2 +4(x+1) (√a)+(√b)=((4b+19c+90)/(5(√c))) 2(√a)= ((25c+4b+19c+90)/(5(√c))) 100ac=(4b+44c+90)^2 100(5c^2 +4c)c={4(c^2 −3c−18) +44c+90}^2 ⇒ 500c^3 +400c^2 =(4c^2 +32c+18)^2 25c^2 (5c+4)=(2c^2 +16c+9)^2 this need to be solved Thereafter as x+1=c x=c−1$
	$5 x^{2} + 14 x + 9 = 5 (x^{2} - x - 20)$ $+ 19 (x + 1) + 90$ $\sqrt{a} - \sqrt{b} = 5 \sqrt{c}$ $a = 5 b + 19 c + 90$ $b = c^{2} - 3 c - 18$ $\Rightarrow a - b = 4 b + 19 c + 90$ $A l s o a = 5 {(x + 1)}^{2} + 4 (x + 1)$ $\sqrt{a} + \sqrt{b} = \frac{4 b + 19 c + 90}{5 \sqrt{c}}$ $2 \sqrt{a} = \frac{25 c + 4 b + 19 c + 90}{5 \sqrt{c}}$ $100 a c = {(4 b + 44 c + 90)}^{2}$ $100 (5 c^{2} + 4 c) c = {4 (c^{2} - 3 c - 18)$ ${+ 44 c + 90}}^{2}$ $\Rightarrow 500 c^{3} + 400 c^{2}$ $= {(4 c^{2} + 32 c + 18)}^{2}$ $25 c^{2} (5 c + 4) = {(2 c^{2} + 16 c + 9)}^{2}$ $t h i s n e e d t o b e s o l v e d$ $T h e r e a f t e r a s x + 1 = c$ $x = c - 1$
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