Question 1171541
If 2 times a number is decreased by 9, the principal square root of this difference is 6 less than the number.
 Find the numbers (s)
:
The way I see this
{{{sqrt(2n-9) = n - 6}}}
square both sides
{{{2n - 9 = (n-6)^2}}}
FOIL (n-6)(n-6)
{{{2n - 9 = n^2 - 12n + 36}}}
combine like terms on the right
{{{0 = n^2 - 12n - 2n + 36 + 9}}}
a quadratic equation
{{{n^2 - 14n + 45 = 0}}}
Factors to
(n-5)(n-9) = 0
Two solutions
n = 5
n = 9
Check both in the original equation
{{{sqrt(2(5)-9) = 5 - 6}}}
{{{sqrt(10-9) = 5 - 6}}}
{{{sqrt(1) = -1}}}
and
{{{sqrt(2(9)-9) = 9 - 6}}}
{{{sqrt(9) = 3}}}
I think the only real solution is n=9