If 1 is a root of the equation ay2+ay+3=0 and y2+y+b=0, then fing ab.
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Answered by
343
If 1 is the root of the given equations then it would satisfy the conditions ;
a(1)^2+a(1)+3 = 0
2a = -3
a = -3/2
(1)^2+(1)+b =0
2 + b = 0
b = -2
ab = (-3/2)×-2
= 3
a(1)^2+a(1)+3 = 0
2a = -3
a = -3/2
(1)^2+(1)+b =0
2 + b = 0
b = -2
ab = (-3/2)×-2
= 3
Answered by
101
Answer:
ab= 3
Step-by-step explanation:
a(1)^2 + a(1) + 3 = 0
a + a + 3 = 0
2a + 3 = 0
2a = -3
a = -3/2
(1)^2 + (1) + b = 0
1 + 1 + b = 0
2 + b = 0
b = -2
Thus, ab= -3/2*2 = 3
( 2 gets cancelled. )
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