mathmath333
  • mathmath333
solve for \(x\)
Mathematics
chestercat
  • chestercat
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mathmath333
  • mathmath333
\(\large \color{black}{\begin{align}|x-6|
phi
  • phi
*IF* x>= 6 , you can drop the absolute values, and (try to) solve \[ x - 6 < x^2 -5x+9 \]
mathmath333
  • mathmath333
\(6\leq x<+\infty\) , for \(x\geq 6 \) is this correct

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phi
  • phi
in other words, solve \[ x^2 -6x+15>0 \] this is a parabola, and you can complete the square to write it as \[ (x-3)^2 + 6 >0 \] because (x-3)^2 is 0 or more, the left side will always be bigger than 0 but we assumed x>=6, so we can say x>=6 is part of the solution
mathmath333
  • mathmath333
\(6\leq x<+\infty\) is this correct
phi
  • phi
yes, that is part of the answer. Now we have to look at the case x<6 then | x - 6| can be written as -(x-6) = -x + 6 (if x is less than or equal to 6)
phi
  • phi
so now we look at \[ -x + 6 < x^2-5x+9\]
phi
  • phi
or \[ x^2 -4x +3 > 0 \] or, factoring \[ (x-1)(x-3)> 0 \] and assuming \( x \le 6\)
phi
  • phi
(x-1)(x-3) is bigger than 0 if it is positive it is positive if both factors are negative (because minus times minus is positive) thus if *both* x-1<0 and x-3 < 0 or x<1 and at the same time x<3 we will get a positive number. x<1 and x<3 simplifies to x<1 thus x<1 (which is consistent with x<=6, our original assumption) is part of the solution
phi
  • phi
(x-1)(x-3) is also positive if both factors are positive. this requires both x>1 and x>3 i..e x>3 thus 3 < x <=6 is part of the solution if we put it together \[ -\infty < x < 1 \text{ or } 3 < x < \infty \]
phi
  • phi
the solution is all real numbers except the closed interval [1,3]
mathmath333
  • mathmath333
thnk u so much
phi
  • phi
yw
phi
  • phi
any questions on the steps?
mathmath333
  • mathmath333
i m reading it

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