Question

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

Answer 1

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

$x_{1}=\frac{-b-\sqrt{\Delta }}{2a} =\frac{8- \sqrt{36}}{2}=\frac{8-6}{2}= \frac{2}{2}= 1 \\ \\x_{2}=\frac{-b+\sqrt{\Delta }}{2a} =\frac{8+ \sqrt{36}}{2}=\frac{8+6}{2}= \frac{14}{2}= 7 \\ \\ Answer : x=0 \ \ \vee \ \ x=1 \ \ \vee \ \ x= 7$

Answer 2

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


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