Question 947657
If you mean {{{sqrt(7x^2-58x+16)}}}=sqrt(7x^2-58x+16),
you were missing some crucial parentheses.
{{{sqrt(7x^2-58x+16)}}} is a real number when {{{7x^2-58x+16>=0}}} .
Otherwise, for example for {{{x=1}}} ,
what's inside the square root is negative,
and the square root is not defined as a real number.
 
{{{y=7x^2-58x+16}}}is a quadratic function that graphs as a parabola.
You can try to find its zeros by using the quadratic formula.
It does have zeros, and they are rational numbers,
so factoring also works in this case.
{{{7x^2-58x+16=(7x-2)(x-8)}}} is zero at {{{x=2/7}}} and at {{{x=8}}} .
In between those numbers, {{{7x-2>0}}} and {{{x-8<0}}} ,
so their product, {{{7x^2-58x+16=(7x-2)(x-8)}}} , is negative,
and {{{sqrt(7x^2-58x+16)}}} is not defined as a real number.
For any other value of {{{x}}} ,
{{{7x^2-58x+16=(7x-2)(x-8)>=0}}} ,
and {{{sqrt(7x^2-58x+16)}}} is defined as a real number.
So the answer is {{{"("}}}{{{-infinity}}}{{{","}}}{{{2/7}}}{{{"] U ["}}}{{{8}}}{{{","}}}{{{infinity}}}{{{")"}}} .