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

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

$${Solve}\:{the}\:{equation}: \\ $$$$\sqrt{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{14}{x}+\mathrm{9}}−\sqrt{{x}^{\mathrm{2}} −{x}−\mathrm{20}}=\mathrm{5}\sqrt{{x}+\mathrm{1}} \\ $$

Answered by Frix last updated on 27/Sep/23

Squaring, transforming, solving, testing ⇒ x_1 =8 x_2 =((5+(√(61)))/2)

$$\mathrm{Squaring},\:\mathrm{transforming},\:\mathrm{solving},\:\mathrm{testing} \\ $$$$\Rightarrow \\ $$$${x}_{\mathrm{1}} =\mathrm{8} \\ $$$${x}_{\mathrm{2}} =\frac{\mathrm{5}+\sqrt{\mathrm{61}}}{\mathrm{2}} \\ $$

Commented by tri26112004 last updated on 27/Sep/23

can you show your solution

$${can}\:{you}\:{show}\:{your}\:{solution} \\ $$

Commented by Frix last updated on 27/Sep/23

(√a)−(√b)=(√c) (√a)=(√b)+(√c) a=b+2(√(bc))+c a−(b+c)=2(√(bc)) (a−(b+c))^2 =4bc a^2 +b^2 +c^2 −2(ab+ac+bc)=0 Now insert a=5x^2 +14x+9 b=x^2 −x−20 c=25(x+1) x^4 −((45x^3 )/4)+((33x^2 )/4)+((505x)/4)+126=0 (x−8)(x+(7/4))(x^2 −5x−9)=0 Test the solutions, some might be false due to squaring

$$\sqrt{{a}}−\sqrt{{b}}=\sqrt{{c}} \\ $$$$\sqrt{{a}}=\sqrt{{b}}+\sqrt{{c}} \\ $$$${a}={b}+\mathrm{2}\sqrt{{bc}}+{c} \\ $$$${a}−\left({b}+{c}\right)=\mathrm{2}\sqrt{{bc}} \\ $$$$\left({a}−\left({b}+{c}\right)\right)^{\mathrm{2}} =\mathrm{4}{bc} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}\left({ab}+{ac}+{bc}\right)=\mathrm{0} \\ $$$$\mathrm{Now}\:\mathrm{insert} \\ $$$${a}=\mathrm{5}{x}^{\mathrm{2}} +\mathrm{14}{x}+\mathrm{9} \\ $$$${b}={x}^{\mathrm{2}} −{x}−\mathrm{20} \\ $$$${c}=\mathrm{25}\left({x}+\mathrm{1}\right) \\ $$$${x}^{\mathrm{4}} −\frac{\mathrm{45}{x}^{\mathrm{3}} }{\mathrm{4}}+\frac{\mathrm{33}{x}^{\mathrm{2}} }{\mathrm{4}}+\frac{\mathrm{505}{x}}{\mathrm{4}}+\mathrm{126}=\mathrm{0} \\ $$$$\left({x}−\mathrm{8}\right)\left({x}+\frac{\mathrm{7}}{\mathrm{4}}\right)\left({x}^{\mathrm{2}} −\mathrm{5}{x}−\mathrm{9}\right)=\mathrm{0} \\ $$$$\mathrm{Test}\:\mathrm{the}\:\mathrm{solutions},\:\mathrm{some}\:\mathrm{might}\:\mathrm{be}\:\mathrm{false}\:\mathrm{due} \\ $$$$\mathrm{to}\:\mathrm{squaring} \\ $$

Commented by tri26112004 last updated on 28/Sep/23

Do you have another solution¿

$${Do}\:{you}\:{have}\:{another}\:{solution}¿ \\ $$

Commented by tri26112004 last updated on 28/Sep/23

Such as Using Inequality

$${Such}\:{as}\:{Using}\:{Inequality} \\ $$

Answered by Rasheed.Sindhi last updated on 28/Sep/23

(√(5x^2 +14x+9))−(√(x^2 −x−20))=5(√(x+1)) x^2 −x−20≥0 ∧ 5x^2 +14x+9≥0 ∧ x+1≥0 (x−5)(x+4)≥0 x−5≥ 0 ∧ x+4≥0 x≥5 ∧ x≥−4⇒x≥5.......(i) 5x^2 +14x+9≥0 (x+1)(5x+9)≥0 x≥−1∧x≥−(9/5)⇒x≥−1....(ii) x+1≥0⇒x≥−1............(iii) (i),(ii) & (iii): x≥5 ∧ x≥−1 ⇒ x≥5 (√(5x^2 +14x+9)) >(√(x^2 −x−20)) 5x^2 +14x+9 >x^2 −x−20 4x^2 +15x+29>0

$$\sqrt{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{14}{x}+\mathrm{9}}−\sqrt{{x}^{\mathrm{2}} −{x}−\mathrm{20}}=\mathrm{5}\sqrt{{x}+\mathrm{1}} \\ $$$${x}^{\mathrm{2}} −{x}−\mathrm{20}\geqslant\mathrm{0}\:\wedge\:\mathrm{5}{x}^{\mathrm{2}} +\mathrm{14}{x}+\mathrm{9}\geqslant\mathrm{0}\:\wedge\:{x}+\mathrm{1}\geqslant\mathrm{0} \\ $$$$\left({x}−\mathrm{5}\right)\left({x}+\mathrm{4}\right)\geqslant\mathrm{0} \\ $$$${x}−\mathrm{5}\geqslant\:\mathrm{0}\:\wedge\:{x}+\mathrm{4}\geqslant\mathrm{0} \\ $$$${x}\geqslant\mathrm{5}\:\wedge\:{x}\geqslant−\mathrm{4}\Rightarrow{x}\geqslant\mathrm{5}.......\left({i}\right) \\ $$$$ \\ $$$$\mathrm{5}{x}^{\mathrm{2}} +\mathrm{14}{x}+\mathrm{9}\geqslant\mathrm{0} \\ $$$$\left({x}+\mathrm{1}\right)\left(\mathrm{5}{x}+\mathrm{9}\right)\geqslant\mathrm{0} \\ $$$${x}\geqslant−\mathrm{1}\wedge{x}\geqslant−\frac{\mathrm{9}}{\mathrm{5}}\Rightarrow{x}\geqslant−\mathrm{1}....\left({ii}\right) \\ $$$${x}+\mathrm{1}\geqslant\mathrm{0}\Rightarrow{x}\geqslant−\mathrm{1}............\left({iii}\right) \\ $$$$ \\ $$$$\left({i}\right),\left({ii}\right)\:\&\:\left({iii}\right): \\ $$$${x}\geqslant\mathrm{5}\:\wedge\:{x}\geqslant−\mathrm{1} \\ $$$$\Rightarrow\:\:\:{x}\geqslant\mathrm{5} \\ $$$$ \\ $$$$\sqrt{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{14}{x}+\mathrm{9}}\:>\sqrt{{x}^{\mathrm{2}} −{x}−\mathrm{20}} \\ $$$$\mathrm{5}{x}^{\mathrm{2}} +\mathrm{14}{x}+\mathrm{9}\:>{x}^{\mathrm{2}} −{x}−\mathrm{20} \\ $$$$\mathrm{4}{x}^{\mathrm{2}} +\mathrm{15}{x}+\mathrm{29}>\mathrm{0} \\ $$$$ \\ $$

Commented by tri26112004 last updated on 28/Sep/23

Can you use the inequality¿

$${Can}\:{you}\:{use}\:{the}\:{inequality}¿ \\ $$

Commented by Rasheed.Sindhi last updated on 28/Sep/23

I′ve tried that but failed to get exact answer.

$${I}'{ve}\:{tried}\:{that}\:{but}\:{failed}\:{to}\:{get}\:{exact}\:{answer}. \\ $$

Commented by tri26112004 last updated on 28/Sep/23

But it has some beauty root x=8...

$${But}\:{it}\:{has}\:{some}\:{beauty}\:{root} \\ $$$${x}=\mathrm{8}... \\ $$

Commented by Frix last updated on 28/Sep/23

The solution method doesn′t care whether you like the solution or not.

$$\mathrm{The}\:\mathrm{solution}\:\mathrm{method}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{care}\:\mathrm{whether} \\ $$$$\mathrm{you}\:\mathrm{like}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{or}\:\mathrm{not}. \\ $$

Commented by tri26112004 last updated on 28/Sep/23

okay

$${okay} \\ $$

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

$$\mathrm{5}{x}^{\mathrm{2}} +\mathrm{14}{x}+\mathrm{9}=\mathrm{5}\left({x}^{\mathrm{2}} −{x}−\mathrm{20}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{19}\left({x}+\mathrm{1}\right)+\mathrm{90} \\ $$$$\sqrt{{a}}−\sqrt{{b}}=\mathrm{5}\sqrt{{c}} \\ $$$${a}=\mathrm{5}{b}+\mathrm{19}{c}+\mathrm{90} \\ $$$${b}={c}^{\mathrm{2}} −\mathrm{3}{c}−\mathrm{18} \\ $$$$\Rightarrow\:\:{a}−{b}=\mathrm{4}{b}+\mathrm{19}{c}+\mathrm{90} \\ $$$${Also}\:\:\:{a}=\mathrm{5}\left({x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{4}\left({x}+\mathrm{1}\right) \\ $$$$\sqrt{{a}}+\sqrt{{b}}=\frac{\mathrm{4}{b}+\mathrm{19}{c}+\mathrm{90}}{\mathrm{5}\sqrt{{c}}} \\ $$$$\mathrm{2}\sqrt{{a}}=\:\frac{\mathrm{25}{c}+\mathrm{4}{b}+\mathrm{19}{c}+\mathrm{90}}{\mathrm{5}\sqrt{{c}}} \\ $$$$\mathrm{100}{ac}=\left(\mathrm{4}{b}+\mathrm{44}{c}+\mathrm{90}\right)^{\mathrm{2}} \\ $$$$\mathrm{100}\left(\mathrm{5}{c}^{\mathrm{2}} +\mathrm{4}{c}\right){c}=\left\{\mathrm{4}\left({c}^{\mathrm{2}} −\mathrm{3}{c}−\mathrm{18}\right)\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{44}{c}+\mathrm{90}\right\}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\mathrm{500}{c}^{\mathrm{3}} +\mathrm{400}{c}^{\mathrm{2}} \\ $$$$\:\:\:\:=\left(\mathrm{4}{c}^{\mathrm{2}} +\mathrm{32}{c}+\mathrm{18}\right)^{\mathrm{2}} \\ $$$$\mathrm{25}{c}^{\mathrm{2}} \left(\mathrm{5}{c}+\mathrm{4}\right)=\left(\mathrm{2}{c}^{\mathrm{2}} +\mathrm{16}{c}+\mathrm{9}\right)^{\mathrm{2}} \\ $$$${this}\:{need}\:{to}\:{be}\:{solved} \\ $$$${Thereafter}\:\:\:\:{as}\:\:\:{x}+\mathrm{1}={c} \\ $$$${x}={c}−\mathrm{1} \\ $$$$ \\ $$

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