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Question Number 223459 by mr W last updated on 26/Jul/25
solve for x∈R (√(25−10x−x^2 ))+(√(15−x^2 ))=2(√5)
$${solve}\:{for}\:{x}\in{R} \\ $$$$\sqrt{\mathrm{25}−\mathrm{10}{x}−{x}^{\mathrm{2}} }+\sqrt{\mathrm{15}−{x}^{\mathrm{2}} }=\mathrm{2}\sqrt{\mathrm{5}} \\ $$
Answered by fantastic last updated on 26/Jul/25
(√(25−10x−x^2 ))=2(√5)−(√(15−x^2 )) or ((√(25−10x−x^2 )))^2 =(2(√5)−(√(15−x^2 )))^2 or 25−10x−x^2 =20+15−x^2 −4(√(75−5x^2 )) or 25−10x=35−4(√(75−5x^2 )) or −4(√(75−5x^2 ))=−10−10x or 2(√(75−5x^2 ))=5+5x or 4(75−5x^2 )=25+25x^2 +50x or 300−20x^2 =25+25x^2 +50x or 60−4x^2 =5+5x^2 +10x or 9x^2 +10x−55=0 So x=((−10±(√(100+1980)))/(18)) =((−10±(√(2080)))/(18))=((−10±(√(16×130)))/(18))=((−5±2(√(130)))/9) ∴ x=((2(√(130))−5)/9) [((−5−2(√(130)))/9) doesnt fullfil the equation ] So x=((2(√(130))−5)/9)≈1.97816761133142
$$\sqrt{\mathrm{25}−\mathrm{10}{x}−{x}^{\mathrm{2}} }=\mathrm{2}\sqrt{\mathrm{5}}−\sqrt{\mathrm{15}−{x}^{\mathrm{2}} } \\ $$$${or}\:\left(\sqrt{\mathrm{25}−\mathrm{10}{x}−{x}^{\mathrm{2}} }\right)^{\mathrm{2}} =\left(\mathrm{2}\sqrt{\mathrm{5}}−\sqrt{\mathrm{15}−{x}^{\mathrm{2}} }\right)^{\mathrm{2}} \\ $$$${or}\:\mathrm{25}−\mathrm{10}{x}−\cancel{{x}^{\mathrm{2}} }=\mathrm{20}+\mathrm{15}−\cancel{{x}^{\mathrm{2}} }−\mathrm{4}\sqrt{\mathrm{75}−\mathrm{5}{x}^{\mathrm{2}} } \\ $$$${or}\:\mathrm{25}−\mathrm{10}{x}=\mathrm{35}−\mathrm{4}\sqrt{\mathrm{75}−\mathrm{5}{x}^{\mathrm{2}} } \\ $$$${or}\:−\mathrm{4}\sqrt{\mathrm{75}−\mathrm{5}{x}^{\mathrm{2}} }=−\mathrm{10}−\mathrm{10}{x} \\ $$$${or}\:\mathrm{2}\sqrt{\mathrm{75}−\mathrm{5}{x}^{\mathrm{2}} }=\mathrm{5}+\mathrm{5}{x} \\ $$$${or}\:\mathrm{4}\left(\mathrm{75}−\mathrm{5}{x}^{\mathrm{2}} \right)=\mathrm{25}+\mathrm{25}{x}^{\mathrm{2}} +\mathrm{50}{x} \\ $$$${or}\:\mathrm{300}−\mathrm{20}{x}^{\mathrm{2}} =\mathrm{25}+\mathrm{25}{x}^{\mathrm{2}} +\mathrm{50}{x} \\ $$$${or}\:\mathrm{60}−\mathrm{4}{x}^{\mathrm{2}} =\mathrm{5}+\mathrm{5}{x}^{\mathrm{2}} +\mathrm{10}{x} \\ $$$${or}\:\mathrm{9}{x}^{\mathrm{2}} +\mathrm{10}{x}−\mathrm{55}=\mathrm{0} \\ $$$${So}\:{x}=\frac{−\mathrm{10}\pm\sqrt{\mathrm{100}+\mathrm{1980}}}{\mathrm{18}} \\ $$$$=\frac{−\mathrm{10}\pm\sqrt{\mathrm{2080}}}{\mathrm{18}}=\frac{−\mathrm{10}\pm\sqrt{\mathrm{16}×\mathrm{130}}}{\mathrm{18}}=\frac{−\mathrm{5}\pm\mathrm{2}\sqrt{\mathrm{130}}}{\mathrm{9}} \\ $$$$\therefore\:{x}=\frac{\mathrm{2}\sqrt{\mathrm{130}}−\mathrm{5}}{\mathrm{9}}\:\left[\frac{−\mathrm{5}−\mathrm{2}\sqrt{\mathrm{130}}}{\mathrm{9}}\:{doesnt}\:{fullfil}\:{the}\:{equation}\:\right] \\ $$$${So}\:{x}=\frac{\mathrm{2}\sqrt{\mathrm{130}}−\mathrm{5}}{\mathrm{9}}\approx\mathrm{1}.\mathrm{97816761133142} \\ $$
Commented by mr W last updated on 26/Jul/25
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Answered by behi834171 last updated on 26/Jul/25
{25−10x−x^2 ≥0}=t^2 ,{15−x^2 ≥0}=s^2 ⇒ { ((t+s=2(√5))),((t^2 −s^2 =10(1−x))) :}⇒ { ((t+s=2(√5))),((t−s=(√5)(1−x))) :} ⇒t=((√5)/2)(3−x),s=((√5)/2)(1+x) ⇒15−x^2 =(5/4)(x^2 +2x+1)⇒ 9x^2 +10x−55=0 x=((−10±(√(100+36×55)))/(18))=((−5±2(√(130)))/9)
$$\left\{\mathrm{25}−\mathrm{10}\boldsymbol{{x}}−\boldsymbol{{x}}^{\mathrm{2}} \geqslant\mathrm{0}\right\}=\boldsymbol{{t}}^{\mathrm{2}} ,\left\{\mathrm{15}−\boldsymbol{{x}}^{\mathrm{2}} \geqslant\mathrm{0}\right\}=\boldsymbol{{s}}^{\mathrm{2}} \\ $$$$\Rightarrow\begin{cases}{\boldsymbol{{t}}+\boldsymbol{{s}}=\mathrm{2}\sqrt{\mathrm{5}}}\\{\boldsymbol{{t}}^{\mathrm{2}} −\boldsymbol{{s}}^{\mathrm{2}} =\mathrm{10}\left(\mathrm{1}−\boldsymbol{{x}}\right)}\end{cases}\Rightarrow\begin{cases}{\boldsymbol{{t}}+\boldsymbol{{s}}=\mathrm{2}\sqrt{\mathrm{5}}}\\{\boldsymbol{{t}}−\boldsymbol{{s}}=\sqrt{\mathrm{5}}\left(\mathrm{1}−\boldsymbol{{x}}\right)}\end{cases} \\ $$$$\Rightarrow{t}=\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\left(\mathrm{3}−{x}\right),{s}=\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\left(\mathrm{1}+{x}\right) \\ $$$$\Rightarrow\mathrm{15}−{x}^{\mathrm{2}} =\frac{\mathrm{5}}{\mathrm{4}}\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}\right)\Rightarrow \\ $$$$\mathrm{9}{x}^{\mathrm{2}} +\mathrm{10}{x}−\mathrm{55}=\mathrm{0} \\ $$$${x}=\frac{−\mathrm{10}\pm\sqrt{\mathrm{100}+\mathrm{36}×\mathrm{55}}}{\mathrm{18}}=\frac{−\mathrm{5}\pm\mathrm{2}\sqrt{\mathrm{130}}}{\mathrm{9}} \\ $$
Commented by mr W last updated on 26/Jul/25
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Answered by mr W last updated on 26/Jul/25
Commented by mr W last updated on 26/Jul/25
(√((5(√2))^2 −(5+x)^2 ))+(√(((√(15)))^2 −x^2 ))=2(√5) the geometric meaning of this equation see diagram. (√((2(√5))^2 +5^2 ))=3(√5) cos β=(((5(√2))^2 +(3(√5))^2 −((√(15)))^2 )/(2(5(√2))(3(√5))))=(8/( 3(√(10)))) 5+x=5(√2)×sin (α+β) =5(√2)(sin α cos β+cos α sin β) =5(√2)((5/(3(√5)))×(8/(3(√(10))))+((2(√5))/(3(√5)))×((√(26))/(3(√(10))))) =((40+2(√(130)))/( 9)) ⇒x=((40+2(√(130)))/( 9))−5=((2(√(130))−5)/9)≈1.978
$$\sqrt{\left(\mathrm{5}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} −\left(\mathrm{5}+{x}\right)^{\mathrm{2}} }+\sqrt{\left(\sqrt{\mathrm{15}}\right)^{\mathrm{2}} −{x}^{\mathrm{2}} }=\mathrm{2}\sqrt{\mathrm{5}} \\ $$$${the}\:{geometric}\:{meaning}\:{of}\:{this} \\ $$$${equation}\:{see}\:{diagram}. \\ $$$$\sqrt{\left(\mathrm{2}\sqrt{\mathrm{5}}\right)^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} }=\mathrm{3}\sqrt{\mathrm{5}} \\ $$$$\mathrm{cos}\:\beta=\frac{\left(\mathrm{5}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} +\left(\mathrm{3}\sqrt{\mathrm{5}}\right)^{\mathrm{2}} −\left(\sqrt{\mathrm{15}}\right)^{\mathrm{2}} }{\mathrm{2}\left(\mathrm{5}\sqrt{\mathrm{2}}\right)\left(\mathrm{3}\sqrt{\mathrm{5}}\right)}=\frac{\mathrm{8}}{\:\mathrm{3}\sqrt{\mathrm{10}}} \\ $$$$\mathrm{5}+{x}=\mathrm{5}\sqrt{\mathrm{2}}×\mathrm{sin}\:\left(\alpha+\beta\right) \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{5}\sqrt{\mathrm{2}}\left(\mathrm{sin}\:\alpha\:\mathrm{cos}\:\beta+\mathrm{cos}\:\alpha\:\mathrm{sin}\:\beta\right) \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{5}\sqrt{\mathrm{2}}\left(\frac{\mathrm{5}}{\mathrm{3}\sqrt{\mathrm{5}}}×\frac{\mathrm{8}}{\mathrm{3}\sqrt{\mathrm{10}}}+\frac{\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{3}\sqrt{\mathrm{5}}}×\frac{\sqrt{\mathrm{26}}}{\mathrm{3}\sqrt{\mathrm{10}}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{40}+\mathrm{2}\sqrt{\mathrm{130}}}{\:\mathrm{9}} \\ $$$$\Rightarrow{x}=\frac{\mathrm{40}+\mathrm{2}\sqrt{\mathrm{130}}}{\:\mathrm{9}}−\mathrm{5}=\frac{\mathrm{2}\sqrt{\mathrm{130}}−\mathrm{5}}{\mathrm{9}}\approx\mathrm{1}.\mathrm{978} \\ $$
Commented by fantastic last updated on 26/Jul/25
Can you find the domain??
$${Can}\:{you}\:{find}\:{the}\:{domain}?? \\ $$
Commented by mr W last updated on 26/Jul/25
−5(√2)≤x+5≤5(√2) ⇒−5((√2)+1)≤x≤5((√2)−1) −(√(15))≤x≤(√(15)) ⇒−(√(15))≤x≤5((√2)−1)≈2.071
$$−\mathrm{5}\sqrt{\mathrm{2}}\leqslant{x}+\mathrm{5}\leqslant\mathrm{5}\sqrt{\mathrm{2}}\:\Rightarrow−\mathrm{5}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)\leqslant{x}\leqslant\mathrm{5}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right) \\ $$$$−\sqrt{\mathrm{15}}\leqslant{x}\leqslant\sqrt{\mathrm{15}} \\ $$$$\Rightarrow−\sqrt{\mathrm{15}}\leqslant{x}\leqslant\mathrm{5}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\approx\mathrm{2}.\mathrm{071} \\ $$
Answered by Raphael254 last updated on 26/Jul/25
(√(25−10x−x^2 )) + (√(15 − x^2 )) = 2(√5) (√(25−10x−x^2 ))= 2(√5) − (√(15 − x^2 )) 25−10x−x^2 = 20 − 4(√5)×(√(15−x^2 )) + 15 − x^2 −x^2 + x^2 − 10x + 25 − 20 − 15 = −4(√5)×(√(15−x^2 )) −10x − 10 = −4(√5)×(√(15−x^2 )) 100x^2 + 200x + 100 = 16×5×(15−x^2 ) 100x^2 + 200x + 100 = 1200 − 80x^2 180x^2 + 200x − 1100 = 0 gcd(180, 200, 1100): 180, 200, 1100∣2 ∗ 90, 100, 550∣2 ∗ 45, 50, 275∣2 45, 25, 275∣3 15, 25, 275∣3 5, 25, 275∣5 ∗ 1, 5, 55∣5 1, 1, 11∣11 1, 1, 1 gcd(180, 200, 1100) = 2×2×5 = 20 180x^2 + 200x − 1100 = 0 ⇒ 9x^2 + 10x − 55 = 0 x = ((−10 ± (√((10)^2 − 4×9×−55)))/(2×9)) x = ((−10 ± (√(100 + 1980)))/(18)) x = ((−10 ± (√(2080)))/(18)) 2080∣2 1040∣2 520∣2 260∣2 130∣2 65∣5 13∣13 1 x = ((−10 ± 4(√(130)))/(18)) x = ((−10 ± 4×≈11,4)/(18)) x = ((−10 ± ≈45,6)/(18)) x_1 = ((−10 + ≈45,6)/(18)) = ((≈35,6)/(18)) ≈1,977... x_2 = ((−10 − ≈45,6)/(18)) = ((−≈55,6)/(18)) ≈−3,088... Testing: (√(25 − 10(2) − (2)^2 )) + (√(15 − (2)^2 )) = (√(25 − 20 − 4)) + (√(15 − 4)) = (√1) + (√(11)) = 1 + ≈3,32 = ≈4,32 ✓ (√(25 − 10(−3) − (−3)^2 )) + (√(15 − (−3)^2 )) = (√(25 + 30 − 9)) + (√(15 − 9)) = (√(46)) + (√6) = ≈6,78 + ≈2,45 = ≈9,23 × So x ∈ R that solves the equation (√(25 − 10x − x^2 )) + (√(15 − x^2 )) = 2(√5), is approximately ((178)/(90)) = 1,977...
$$ \\ $$$$\sqrt{\mathrm{25}−\mathrm{10}{x}−{x}^{\mathrm{2}} }\:+\:\sqrt{\mathrm{15}\:−\:{x}^{\mathrm{2}} }\:=\:\mathrm{2}\sqrt{\mathrm{5}} \\ $$$$ \\ $$$$\sqrt{\mathrm{25}−\mathrm{10}{x}−{x}^{\mathrm{2}} \:}=\:\mathrm{2}\sqrt{\mathrm{5}}\:−\:\sqrt{\mathrm{15}\:−\:{x}^{\mathrm{2}} } \\ $$$$\mathrm{25}−\mathrm{10}{x}−{x}^{\mathrm{2}} \:=\:\mathrm{20}\:−\:\mathrm{4}\sqrt{\mathrm{5}}×\sqrt{\mathrm{15}−{x}^{\mathrm{2}} }\:+\:\mathrm{15}\:−\:{x}^{\mathrm{2}} \\ $$$$−{x}^{\mathrm{2}} \:+\:{x}^{\mathrm{2}} \:−\:\mathrm{10}{x}\:+\:\mathrm{25}\:−\:\mathrm{20}\:−\:\mathrm{15}\:=\:−\mathrm{4}\sqrt{\mathrm{5}}×\sqrt{\mathrm{15}−{x}^{\mathrm{2}} } \\ $$$$−\mathrm{10}{x}\:−\:\mathrm{10}\:=\:−\mathrm{4}\sqrt{\mathrm{5}}×\sqrt{\mathrm{15}−{x}^{\mathrm{2}} } \\ $$$$\mathrm{100}{x}^{\mathrm{2}} \:+\:\mathrm{200}{x}\:+\:\mathrm{100}\:=\:\mathrm{16}×\mathrm{5}×\left(\mathrm{15}−{x}^{\mathrm{2}} \right) \\ $$$$\mathrm{100}{x}^{\mathrm{2}} \:+\:\mathrm{200}{x}\:+\:\mathrm{100}\:=\:\mathrm{1200}\:−\:\mathrm{80}{x}^{\mathrm{2}} \\ $$$$\mathrm{180}{x}^{\mathrm{2}} \:+\:\mathrm{200}{x}\:−\:\mathrm{1100}\:=\:\mathrm{0} \\ $$$$ \\ $$$${gcd}\left(\mathrm{180},\:\mathrm{200},\:\mathrm{1100}\right): \\ $$$$ \\ $$$$\mathrm{180},\:\mathrm{200},\:\mathrm{1100}\mid\mathrm{2}\:\ast \\ $$$$\mathrm{90},\:\mathrm{100},\:\mathrm{550}\mid\mathrm{2}\:\ast \\ $$$$\mathrm{45},\:\mathrm{50},\:\mathrm{275}\mid\mathrm{2} \\ $$$$\mathrm{45},\:\mathrm{25},\:\mathrm{275}\mid\mathrm{3} \\ $$$$\mathrm{15},\:\mathrm{25},\:\mathrm{275}\mid\mathrm{3} \\ $$$$\mathrm{5},\:\mathrm{25},\:\mathrm{275}\mid\mathrm{5}\:\ast \\ $$$$\mathrm{1},\:\mathrm{5},\:\mathrm{55}\mid\mathrm{5} \\ $$$$\mathrm{1},\:\mathrm{1},\:\mathrm{11}\mid\mathrm{11} \\ $$$$\mathrm{1},\:\mathrm{1},\:\mathrm{1} \\ $$$$ \\ $$$${gcd}\left(\mathrm{180},\:\mathrm{200},\:\mathrm{1100}\right)\:=\:\mathrm{2}×\mathrm{2}×\mathrm{5}\:=\:\mathrm{20} \\ $$$$ \\ $$$$\mathrm{180}{x}^{\mathrm{2}} \:+\:\mathrm{200}{x}\:−\:\mathrm{1100}\:=\:\mathrm{0}\:\Rightarrow\:\mathrm{9}{x}^{\mathrm{2}} \:+\:\mathrm{10}{x}\:−\:\mathrm{55}\:=\:\mathrm{0} \\ $$$$ \\ $$$${x}\:=\:\frac{−\mathrm{10}\:\pm\:\sqrt{\left(\mathrm{10}\right)^{\mathrm{2}} \:−\:\mathrm{4}×\mathrm{9}×−\mathrm{55}}}{\mathrm{2}×\mathrm{9}} \\ $$$${x}\:=\:\frac{−\mathrm{10}\:\pm\:\sqrt{\mathrm{100}\:+\:\mathrm{1980}}}{\mathrm{18}} \\ $$$${x}\:=\:\frac{−\mathrm{10}\:\pm\:\sqrt{\mathrm{2080}}}{\mathrm{18}} \\ $$$$ \\ $$$$\mathrm{2080}\mid\mathrm{2} \\ $$$$\mathrm{1040}\mid\mathrm{2} \\ $$$$\mathrm{520}\mid\mathrm{2} \\ $$$$\mathrm{260}\mid\mathrm{2} \\ $$$$\mathrm{130}\mid\mathrm{2} \\ $$$$\mathrm{65}\mid\mathrm{5} \\ $$$$\mathrm{13}\mid\mathrm{13} \\ $$$$\mathrm{1} \\ $$$$ \\ $$$${x}\:=\:\frac{−\mathrm{10}\:\pm\:\mathrm{4}\sqrt{\mathrm{130}}}{\mathrm{18}} \\ $$$${x}\:=\:\frac{−\mathrm{10}\:\pm\:\mathrm{4}×\approx\mathrm{11},\mathrm{4}}{\mathrm{18}} \\ $$$${x}\:=\:\frac{−\mathrm{10}\:\pm\:\approx\mathrm{45},\mathrm{6}}{\mathrm{18}} \\ $$$$ \\ $$$${x}_{\mathrm{1}} \:=\:\frac{−\mathrm{10}\:+\:\approx\mathrm{45},\mathrm{6}}{\mathrm{18}}\:=\:\frac{\approx\mathrm{35},\mathrm{6}}{\mathrm{18}}\:\approx\mathrm{1},\mathrm{977}… \\ $$$${x}_{\mathrm{2}} \:=\:\frac{−\mathrm{10}\:−\:\approx\mathrm{45},\mathrm{6}}{\mathrm{18}}\:=\:\frac{−\approx\mathrm{55},\mathrm{6}}{\mathrm{18}}\:\approx−\mathrm{3},\mathrm{088}… \\ $$$$ \\ $$$${Testing}: \\ $$$$ \\ $$$$\sqrt{\mathrm{25}\:−\:\mathrm{10}\left(\mathrm{2}\right)\:−\:\left(\mathrm{2}\right)^{\mathrm{2}} }\:+\:\sqrt{\mathrm{15}\:−\:\left(\mathrm{2}\right)^{\mathrm{2}} }\:=\:\sqrt{\mathrm{25}\:−\:\mathrm{20}\:−\:\mathrm{4}}\:+\:\sqrt{\mathrm{15}\:−\:\mathrm{4}}\:=\:\sqrt{\mathrm{1}}\:+\:\sqrt{\mathrm{11}}\:=\:\mathrm{1}\:+\:\approx\mathrm{3},\mathrm{32}\:=\:\approx\mathrm{4},\mathrm{32}\:\checkmark \\ $$$$\sqrt{\mathrm{25}\:−\:\mathrm{10}\left(−\mathrm{3}\right)\:−\:\left(−\mathrm{3}\right)^{\mathrm{2}} }\:+\:\sqrt{\mathrm{15}\:−\:\left(−\mathrm{3}\right)^{\mathrm{2}} }\:=\:\sqrt{\mathrm{25}\:+\:\mathrm{30}\:−\:\mathrm{9}}\:\:+\:\sqrt{\mathrm{15}\:−\:\mathrm{9}}\:=\:\sqrt{\mathrm{46}}\:+\:\sqrt{\mathrm{6}}\:=\:\approx\mathrm{6},\mathrm{78}\:+\:\approx\mathrm{2},\mathrm{45}\:=\:\approx\mathrm{9},\mathrm{23}\:× \\ $$$$ \\ $$$${So}\:{x}\:\in\:\mathbb{R}\:{that}\:{solves}\:{the}\:{equation}\:\sqrt{\mathrm{25}\:−\:\mathrm{10}{x}\:−\:{x}^{\mathrm{2}} }\:+\:\sqrt{\mathrm{15}\:−\:{x}^{\mathrm{2}} }\:=\:\mathrm{2}\sqrt{\mathrm{5}},\:{is}\:{approximately}\:\frac{\mathrm{178}}{\mathrm{90}}\:=\:\mathrm{1},\mathrm{977}… \\ $$

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