+118653 \(\sqrt{3x^2-5x}=\sqrt{x^2+2x+5}\) помогите
Firstly both expressions under the sqrts must be positive. I'll think about that more later.
\(3x^2-5x=x^2+2x+5\\ 2x^2-7x-5=0\\ 2x^2-7x-5=0\\ \triangle=49+40=89\\ x=\frac{7\pm\sqrt{89}}{4}\)
(7+sqrt(89))/4 = 4.108495283014151
(7-sqrt(89))/4 = -0.608495283014151
x=\frac{7\pm\sqrt{89}}{4}
3*(4.108495283014151)^2-5(4.108495283014151) = 30.096724056577831167 >0 The plus version is good
3*(-0.608495283014151)^2-5(-0.608495283014151) = 4.153275943422170167 >0 Good
By visual inspection I can see that the DH sides will be real.
So the answers are \(x=\frac{7\pm\sqrt{89}}{4}\)
Here is the graph
