## mathmath333 one year ago solve for $$x$$

1. mathmath333

\large \color{black}{\begin{align}|x-6|<x^2-5x+9,\ x\in \mathbb{R}\hspace{.33em}\\~\\ \end{align}}

2. phi

*IF* x>= 6 , you can drop the absolute values, and (try to) solve $x - 6 < x^2 -5x+9$

3. mathmath333

$$6\leq x<+\infty$$ , for $$x\geq 6$$ is this correct

4. 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

5. mathmath333

$$6\leq x<+\infty$$ is this correct

6. 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)

7. phi

so now we look at $-x + 6 < x^2-5x+9$

8. phi

or $x^2 -4x +3 > 0$ or, factoring $(x-1)(x-3)> 0$ and assuming $$x \le 6$$

9. 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

10. 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 $-\infty < x < 1 \text{ or } 3 < x < \infty$

11. phi

the solution is all real numbers except the closed interval [1,3]

12. mathmath333

thnk u so much

13. phi

yw

14. phi

any questions on the steps?

15. mathmath333