Question

In the equation above. K is a constant. If x=9, what value is k? A) 1 B) 7 C) 16 D) 79

Answer 1

Answer:

D) k = 79

Step-by-step explanation:

Given equation: $\sf\sqrt{k+2}-x=0$

Steps:

1. Substitute 9 as the value of x in the equation:

$\sf\sqrt{k+2}-9=0$

2. Add 9 to both sides:

$\sf\sqrt{k+2}-9+9=0+9\\\\\Rightarrow \sqrt{k+2}=9$

3. Square both sides of the equation:

$\sf\left(\sqrt{k+2}\right)^2=9^2\\\\\Rightarrow k+2=81$

4. Subtract 2 from both sides:

$\sf k+2-2=81-2\\\\\Rightarrow k = 79$

5. Check your work:

$\sf\sqrt{k+2}-x=0\ \textsf{[ substitute 79 for k, and 9 for x ]}\\\\\sqrt{79+2}-9=0\ \textsf{[ add ]}\\\\\sqrt{81}-9=0\ \textsf{[ take the square root ]}\\\\9-9=0\ \textsf{[ subtract ]}\\\\0=0\ \checkmark$

Answer 2

Answer: 79

Step-by-step explanation:

$\sqrt{k+2}-9=0\\\\\sqrt{k+2}=9\\\\k+2=81\\\\k=\boxed{79}$