Question

I. $12x^2-x-1=0$
II. $20y^2-41y+20=0$

• A. x > y
• B. $x\ge y$
• C. x < y
• D. $x\le y$
• E. Relationship between x and y cannot be determined

Answer 1

The correct option is C x < y

I. $12x^2$ - x - 1 = 0
$x_1=\frac{-b+\sqrt{D}}{2a}=\frac{1+\sqrt{1-4\times 12\times -1}}{24}$
$=\frac{1+7}{24}=\frac{8}{24}=\frac{1}{3}{x}_2=\frac{-b-\sqrt{D}}{2a}{x}_2=\frac{1-7}{24}=\frac{-6}{24}=\frac{-1}{4}x=\frac{1}{3},-\frac{1}{4}$
II. $20y^2$ - 41y + 20
$y_1=\frac{41-\sqrt{1681-1600}}{40}{y}_2=\frac{41-\sqrt{1681-1600}}{40}{y}_1=\frac{41+9}{40}=\frac{50}{40},y_2=\frac{32}{40}y=\frac{5}{4},\frac{4}{5}\Rightarrow x<y$