Question

I. 4$x^2$ = 49
II. 9$y^2$ - 66y + 121 = 0

• A. if x < y
• B. if x y
• C. if x = y
• D. if x y

Answer 1

The correct option is A

if x < y

$4x^2=49$

$x^2=\frac{49}{4}$

$x=\pm \frac{7}{2}=\pm 3.5$

$9y^2-66y+121=0$

$(3y-11{)}^2=0$

$y=\frac{11}{3}=3.66$

x < y