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Find the root of the problem 

√5x^2 - (8√5 +10)x + (20√5 +40) = 0

 

 

Thank you for those who answered!laugh

 Jun 1, 2019
 #1
avatar+9129 
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√5x^2 - (8√5 +10)x + (20√5 +40) = 0

 

\(\sqrt{5}\cdot x^2-(8\cdot \sqrt{5}+10)\cdot x+(20\cdot \sqrt{5}+40)=0\\ \sqrt{5}=w\\ wx^2-(8w+10)\cdot x+(20w+40)=0\)

a                   b                            c

\(x = {8w+10 \pm \sqrt{(8w+10)^2-4(w)(20w+40)} \over 2w}\\ x=\frac{8w+10\pm\sqrt{64w^2+160w+100-80w^2-160w}}{2w}\\ x=\frac{8w+10\pm\sqrt{100-16w^2}}{2w}\\ x=\frac{8w+10\pm\sqrt{(10+4w)\cdot(10-4w)}}{2w}\\ \ w=\sqrt{5}\\ x=\frac{27,88854\pm\sqrt{20}}{4,472136}\)

\(x_1=7,23607\\ x_2=5,23607\)

laugh  !

 Jun 1, 2019

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