Question

(х²-8х+7)(х²-8х+15)=105 решите методом введения новой переменной​

Answer 1

Ответ:

$4+i\sqrt{6}, \quad 4-i\sqrt{6}, \quad 0, \quad 8;$

Объяснение:

$(x^{2}-8x+7)(x^{2}-8x+15)=105;$

$t=x^{2}-8x+7 \Rightarrow x^{2}-8x+15=t+8;$

$t(t+8)=105;$

$t^{2}+8t-105=0;$

$\left \{ {{t_{1}+t_{2}=-8} \atop {t_{1} \cdot t_{2}=-105}} \right. \Leftrightarrow \left \{ {{t_{1}=-15} \atop {t_{2}=7}} \right. ;$

$x^{2}-8x+7=-15 \quad \vee \quad x^{2}-8x+7=7;$

$x^{2}-8x+22=0 \quad \vee \quad x^{2}-8x=0;$

$x^{2}-8x+22=0 \quad \vee \quad x(x-8)=0;$

$x^{2}-8x+22=0 \quad \vee \quad x=0 \quad \vee \quad x-8=0;$

$x^{2}-8x+22=0 \quad \vee \quad x=0 \quad \vee \quad x=8;$

$x^{2}-8x+22=0;$

$D=b^{2}-4ac;$

$D=(-8)^{2}-4 \cdot 1 \cdot 22=64-88=-24;$

$x_{1,2}=\frac{-b \pm \sqrt{D}}{2a};$

$x_{1}=\frac{-(-8)+\sqrt{-24}}{2 \cdot 1}=\frac{8+2i\sqrt{6}}{2}=\frac{8}{2}+\frac{2i\sqrt{6}}{2}=4+i\sqrt{6};$

$x_{2}=\frac{-(-8)-\sqrt{-24}}{2 \cdot 1}=\frac{8-2i\sqrt{6}}{2}=\frac{8}{2}-\frac{2i\sqrt{6}}{2}=4-i\sqrt{6};$