Question

розв'яжіть: 6 + x = $\sqrt{x + 78}$

Answer 1

Ответ:

х=3

Объяснение:

6+х=√(х+78)

(6+х)²=√(х+78)²

36+12х+х²=х+78

х²+11х-42=0; по теореме Виета х=-14 (не подходит)  х=3

Answer 2

$\displaystyle\bf\\6+x=\sqrt{x+78} \\\\ODZ : 6+x\geq 0 \ \ \ \Rightarrow \ \ \ x\geq -6\\\\(6+x)^{2} =(\sqrt{x+78} )^{2} \\\\36+12x+x^{2} =x+78\\\\x^{2} +12x-x+36-78=0\\\\x^{2} +11x-42=0\\\\D=11^{2} -4\cdot(-42)=121+168=289=17^{2}\\\\\\x_{1} =\frac{-11-17}{2} =\frac{-28}{2} =-14\\\\x_{2} =\frac{-11+17}{2} =\frac{6}{2} =3\\\\\\Otvet \ : \ 3$