Question 319415
find a number that is twice its square root,increased by 7 is 9
:
{{{2sqrt(x)}}} + 7 = 9
:
{{{2sqrt(x)}}} = 9 - 7
{{{2sqrt(x)}}} = 2
square both sides
4x = 4
x = 1
:
Check solution in original equation
{{{2sqrt(1)}}} + 7 = 9
2(1) + 7 = 9