Question

If 1 is a root of the equation \[a{{y}^{2}}+ay+3=0\] and \[{{y}^{2}}+y+b=0,\] then ab equals:
\[\left( a \right)3\]
\[\left( b \right)\dfrac{-7}{2}\]
\[\left( c \right)6\]
\[\left( d \right)-3\]

Answer 1

Hint: We are given that 1 is a root of \[a{{y}^{2}}+ay+3=0\] and \[{{y}^{2}}+y+b=0.\] So, we put y = 1 in both the sides and solve for the value of a and for the value of b, after solving when we get the value of a and b, we do the product of a with b to find the value of ab.

Complete step-by-step answer:
We are given the equation \[a{{y}^{2}}+ay+3=0\] and \[{{y}^{2}}+y+b=0.\] It is mentioned that 1 is zero of both the equation. As 1 is zero of both the equations, it means that it must be satisfying both the equations.
So, put y = 1 in \[{{y}^{2}}+y+b=0.\] We get,
\[{{1}^{2}}+1+b=0\]
\[\Rightarrow 1+1+b=0\]
Solving for b, we get,
\[\Rightarrow 2+b=0\]
\[\Rightarrow b=-2\]
So, we get b as – 2.
Now, we put y = 1 in \[a{{y}^{2}}+ay+3=0.\] We get,
\[a{{\left( 1 \right)}^{2}}+a\left( 1 \right)+3=0\]
Simplifying, we get,
\[\Rightarrow a+a+3=0\]
Solving for a , we will get,
\[\Rightarrow 2a+3=0\]
Simplifying further, we get,
\[\Rightarrow 2a=-3\]
Dividing both the sides by 2, we will get,
\[\Rightarrow \dfrac{2a}{2}=\dfrac{-3}{2}\]
\[\Rightarrow a=\dfrac{-3}{2}\]
So, we have a as \[\dfrac{-3}{2}.\]
Now, we have to find the value of the product of a and b. As we have a as \[\dfrac{-3}{2}\] and b as – 2. So,
\[ab=\left( \dfrac{-3}{2} \right)\left( -2 \right)\]
Simplifying, we get,
\[\Rightarrow ab=\dfrac{-3}{2}\times -2\]
\[\Rightarrow ab=-3\times -1\]
\[\Rightarrow ab=3\]
So, we get the product of a and b as 3.

So, the correct answer is “Option (a)”.

Note: The simple mistake that students can possibly make is of \[{{1}^{2}}=2\] which is wrong as \[{{1}^{2}}\] mean the product of 1 with itself.
\[\Rightarrow {{1}^{2}}=1\times 1=1\]
So, while performing this calculation, students have to be careful. Also, the product of two negative numbers will always give us a positive number. So, \[\left( \dfrac{-3}{2} \right)\times \left( -2 \right)=+3\] or 3.