Question

Question 3 Unsaved Solve for x, given the equation Square root of x minus 5 + 7 = 11

Answer 1

$Domain:\\\\x-5\geq0\to x\geq5\\\\\sqrt{x-5}+7=11\ \ \ \ \ |-7\\\\\sqrt{x-5}=4\ \ \ \ \ |^2\\\\(\sqrt{x-5})^2=4^2\to x-5=16\ \ \ \ \ |+5\\\\\boxed{x=21}\in D$

Answer 2

We have the following equation:

$\sqrt{x-5} +7=11$

To solve this, we first need to subtract the 7 on both sides of the equation.

$\sqrt{x-5} +7-7=11-7$

The equation then becomes:

$\sqrt{x-5} =4$

Now we must square both sides of the equation to get rid of the radical/square root symbol.

$(\sqrt{x-5})^2 =(4)^2$

Now we have a normal one step equation.

$x-5=16$

Add 5 on both sides to find x.

$x-5+5=16+5$

x=21

To see if this solution is true, we must substitute 21 back into the original equation.

$\sqrt{21-5} +7=11$

$\sqrt{16} +7=11$

$4+7=11$

11=11

Therefore, the solution is true.