Question

If $6a^2 + 5a + 4 = 3,$ then what is the smallest possible value of $2a + 1$?

Answer 1

Answer: 0

Explanation:

The given equation:
`6a^2+5a+4=3`

Subtract 3 from both the sides, we get

`6a^2+5a+1=0`

Now , we can split 5a as 2a+3a and
`2a* 3a = 6a^2`

So,
`6a^2+5a+1=0\Rightarrow\ 6a^2+2a+3a+1=0`

`\Rightarrow\ 2a(3a+1)+(3a+1)=0\\\\\Rightarrow\ (3a+1)(2a+1)=0\\\\\Rightarrow\ (3a+1)=0\text{ or }(2a+1)=0\\\\\Rightarrow\ a=-(1)/(3)\text{ or }a=-(1)/(2)`

At
`a=-(1)/(3)`

`2a+1=2(-(1)/(3))+1=-(2)/(3)+1=(-2+3)/(3)=\frac{1}3{}`

At
`a=-(1)/(2)`

`2a+1=2(-(1)/(2))+1=-1+1=0`

Since,
`0< (1)/(3)`

Hence, the possible value of 2a+1 is 0.