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Question : 121 of 209
Marks:
+1,
-0
Solution:
(4)I.x2+5x+6=0
⇒x2+2x+3x+6=0
⇒x(x+2)+3(x+2)=0
⇒(x+3)(x+2)=0
⇒x=–3or–2
II.y2+3y+2=0
⇒y2+2y+y+2=0
⇒y(y+2)+1(y+2)=0
⇒(y+1)(y+2)=0
⇒y=–1or–2
$⇒x < y$
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