Question

How can i solve x^3+7x=8x^2

Answer 1

`x^3+7x=8x^2\\ \\x^3-8x^2+7x =0\\ \\x(x^2-8x+7)=0\\ \\x=0 \ \ \vee \ \ x^2-8x+7 =0\\ \\ \Delta = b^(2)-4ac =(-8)^(2)-4*1*7=64 - 28 = 36`


`x_(1)=(-b-√(\Delta ))/(2a) =(8- √(36))/(2)=(8-6)/(2)= (2)/(2)= 1 \\ \\x_(2)=(-b+√(\Delta ))/(2a) =(8+ √(36))/(2)=(8+6)/(2)= (14)/(2)= 7 \\ \\ Answer : x=0 \ \ \vee \ \ x=1 \ \ \vee \ \ x= 7`

Answer 2

`x^3+7x=8x^2\ \ \ /-8x^2\\\\x^3-8x^2+7x=0\\\\x(x^2-8x+7)=0\iff x=0\ \vee\ x^2-8x+7=0`



`x^2-8x+7=0\\\\a=1;\ b=-8;\ c=7\\\\\Delta=b^2-4ac;\ x_1=(-b-\sqrt\Delta)/(2a);\ x_2=(-b+\sqrt\Delta)/(2a)\\\\\Delta=(-8)^2-4\cdot1\cdot7=64-28=36;\ \sqrt\Delta=√(36)=6\\\\x_1=(8-6)/(2\cdot1)=(2)/(2)=1;\ x_2=(8+6)/(2\cdot1)=(14)/(2)=7\\\\\\\\Answer:x=0\ or\ x=1\ or\ x=7.`