What real number is exactly 3 greater than its square root. Write the number to the nearest hundredth.

thank you for answering

Equation is: sqrt(x) = x - 3

--> x = x^2 -6x + 9

--> 0 = x^2 -7x + 9

Use quadratic formula where a=1 , b=-7, c=9 :

x = (7±sqrt(49-4(1)(9)) / 2 = (7±sqrt(13)) / 2

x = 1.70 or x = 5.30

is right, but consider that 1.70 is a false root.

sqrt(1.70)=1.30

1.70-1.30<3.0

Yeah it depends --> Sqrt(1.7) = ±1.3

Works for the negative version: -1.3 + 3 = 1.7

Falls apart for the positive version: 1.3 + 3 =4.3

You can always choose to take the negative version of the squareroot thats why I think x=1.70 is a suitable answer.