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Question
If y = 1 is a common root of the equations \[a y^2 + ay + 3 = 0 \text { and } y^2 + y + b = 0\], then ab equals
Options
3
-7/2
6
-3
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Solution
Since, y = 1 is a root of the equations \[a y^2 + ay + 3 = 0\].
\[\therefore a \left( 1 \right)^2 + a\left( 1 \right) + 3 = 0\]
\[ \Rightarrow 2a + 3 = 0\]
\[ \Rightarrow a = - \frac{3}{2} . . . (1)\]
Since, y = 1 is a root of the equations \[y^2 + y + b = 0\].
So, it satisfies the given equation.
\[\therefore \left( 1 \right)^2 + \left( 1 \right) + b = 0\]
\[ \Rightarrow 2 + b = 0\]
\[ \Rightarrow b = - 2 . . . (2)\]
From (1) and (2),
\[ab = \left( - \frac{3}{2} \right)\left( - 2 \right)\]
\[ = 3\]
Thus, ab is equal to 3.
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