Question

Решите уравнение.
4(x-4)(x+8)=(3x+2)(x-5)+(x-1)

Answer 1

Объяснение:

$\[\begin{gathered}4\left({x-4}\right)\left({x+8}\right)=\left({3x+2}\right)\left({x-5}\right)+\left({x-1}\right)\hfill\\4{x^2}+16x-128=3{x^2}-13x-10+x-1\hfill\\4{x^2}+16x-128-3{x^2}+13x+10-x+1=0\hfill\\\end{gathered}\]$

$\[\begin{gathered}{x^2}+28x-117=0\hfill \\D={b^2}-4ac={28^2}-4\cdot 1 \cdot (-117)=784+468=1252\hfill \\ \end{gathered}\]$

$\[\begin{gathered}{x_{1;2}}=\frac{{-b\pm\sqrt D}}{{2a}}=\frac{{-28\pm\sqrt {1252}}}{{2\cdot 1}}=\frac{{-28\pm2\sqrt{313}}}{2}=\frac{{2\cdot\left({-14\pm\sqrt{313}}\right)}}{2}=-14\pm\sqrt{313}\hfill\\{x_1}=-14+\sqrt{313}\hfill\\{x_2}=-14-\sqrt{313}\hfill\\\end{gathered}\]$