Answer:
1. A. and C.
2. C.
Step-by-step explanation:
[tex]1.\\\bold{YES}\\A.\ (x+5)^2=49\iff x+5=\pm\sqrt{49}\\\\x+5=\pm7\qquad\text{subtract 5 from both sides}\\\\x=-13\in\mathbb{Q}\ or\ x=2\in\mathbb{Q}\\\\C.\ x^2=81\iff x=\pm\sqrt{81}\\\\x=\pm9\in\mathbb{Q}\\\\\bold{NOT}\\B.\ x^2=12\to x=\pm\sqrt{12}\notin\mathbb{Q}\\D.\ (x-1)^2=20\to x-1=\pm\sqrt{20}\notin\mathbb{Q}[/tex]
[tex]\bold{YES}\\C.\\x^2+12x+36=0\\\\x^2+2(x)(6)+6^2=0\qquad\text{use}\ (a+b)^2=a^2+2ab+b^2\\\\(x+6)^2=0\iff x+6=0\qquad\text{subtract 6 from both sides}\\\\x=-6\\\\\bold{NOT}\\A.\\3x^2+x-2=0\\3x^2+3x-2x-2=0\\3x(x+1)-2(x+1)=0\\(x+1)(3x-2)=0\iff x+1=0\ \vee\ 3x-2=0\\\text{two different solutions}\\B.\\x^2-15x+36=0\\x^2-12x-3x+36=0\\x(x-12)-3(x-12)=0\\(x-12)(x-3)=0\iff x-12=0\ \vee\ x-3=0\\\text{two different solutions}[/tex]
