Question

solve for \(x\)

Answer 1

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

Answer 2

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

Answer 3

\(6\leq x<+\infty\) , for \(x\geq 6 \)

is this correct

Answer 4

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

Answer 5

\(6\leq x<+\infty\) is this correct

Answer 6

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)

Answer 7

so now we look at
\[ -x + 6 < x^2-5x+9\]

Answer 8

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

Answer 9

(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

Answer 10

(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
\[ -\infty < x < 1 \text{ or } 3 < x < \infty \]

Answer 11

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

Answer 12

thnk u so much

Answer 13

yw

Answer 14

any questions on the steps?

Answer 15

i m reading it