© examsiri.com
Question : 128 of 200
Marks:
+1,
-0
Solution:
(5)
+x+=0⇒x2+5x+4=0⇒x2+4x+x+4=0⇒x(x+4)+1(x+4)=0⇒(x+1)(x+4)=0⇒x=–1or–4Again,
⇒
3y2+4y+1=0⇒3y2+3y+y+1=0⇒3y(y+1)+1(y+1)=0⇒(y+1)(3y+1)=0⇒y=–1or–Clearly, quantity I
< quantityII
© examsiri.com
Go to Question: