3x2+8x+4=0
4y2−19y+12=0
(a)
(b)
(c)
(d)
(e)
On solving we get x= - 23 or -2
y= 34 or 4
So, clearly, x<y
3x2 + 8x + 4 = 0
4y2 - 19y + 12 = 0
I.x2−7x+12=0 II.y2−12y+32=0
x2 - 1 = 0
y2 + 4y + 3 = 0
I.3x2+8x+4=0
II.4y2−19y+12=0
I. 3x2+ 8x + 4 = 0
II. 4y2 - 19y + 12 = 0