# How close to 0 do we need to take x so that(x^{2}+6x+9)<9.001?

How close to 0 do we need to take x so that
$\left({x}^{2}+6x+9\right)<9.001?$

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Clara Reese

Consider expression,
${x}^{2}+6x+9$

$⇒\left({x}^{2}+3x\right)+\left(3x+9\right)$

$⇒\left(x+3{\right)}^{2}$
Hence the inequality will be,
$\left(x+3{\right)}^{2}<9.001$

Case(i):

Case(ii): $x+3<-3.00016666$

Since we need the solution which is close to zero,
The solution that satisfies should be
$x<0.00016666$
And it can be close to zero by 4 decimals(4 negative powers of 10)