Question

1. Which of the following equations have rational solutions?

Select TWO that apply.

A. (X+5)^2=49
B. X^2=12
C. X^2=81
D. (X-1)^2=20

2. Which quadratic equation is a perfect square trinomial with only one root, x=-6?

A. 3x^2+x-2=0
B. X^2-15x+36=0
C. X^2+12x+36=0

Answer 1

1. A. and C.

2. C.

Explanation:

`1.\\\bold{YES}\\A.\ (x+5)^2=49\iff x+5=\pm√(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√(81)\\\\x=\pm9\in\mathbb{Q}\\\\\bold{NOT}\\B.\ x^2=12\to x=\pm√(12)\\otin\mathbb{Q}\\D.\ (x-1)^2=20\to x-1=\pm√(20)\\otin\mathbb{Q}`

`\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}`