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## Homework Statement

The left side in the equation [itex]\sqrt{3-x}[/itex][itex]\sqrt{x+9}[/itex]=8x+1 is only definable when x is located in a certain interval a [itex]\leq[/itex] x [itex]\leq[/itex] b. What's a & b? (Note: The answer ought to be integers)

## The Attempt at a Solution

[itex]\sqrt{3-x}[/itex][itex]\sqrt{x+9}[/itex]=8x+1

([itex]\sqrt{3-x}[/itex][itex]\sqrt{x+9}[/itex])

^{2}=(8x+1)

^{2}

Which should lead to this:

(3-x)*2([itex]\sqrt{3-x}[/itex][itex]\sqrt{x+9}[/itex])(x+9)=64x

^{2}+22x-26

The problem now is, what shall I do next? Expand the parentheses?

Well.. If I do that then what? I'll obviously get a quadratic equation. Any advise what to do with the left side? Those roots are hopeless.