Answer:
[tex]k=4[/tex]
Step-by-step explanation:
So we have the equation:
[tex]\sqrt{2k^2+17}-x=0[/tex]
Where:
[tex]k>0\text{ and } x=7[/tex]
And we want to find k.
Firstly, let's substitute 7 for x. So:
[tex]\sqrt{2k^2+17}-7=0[/tex]
Now, we can solve for k. Add 7 to both sides:
[tex]\sqrt{2k^2+17}=7[/tex]
Square both sides:
[tex]2k^2+17=49[/tex]
Subtract 17 from both sides:
[tex]2k^2=32[/tex]
Divide both sides by 2:
[tex]k^2=16[/tex]
Take the square root of both sides:
[tex]k=\pm\sqrt{16}[/tex]
Evaluate:
[tex]k=\pm 4[/tex]
Remember that we are told k>0. In other words, k must be positive. So, we can ignore the negative answer, giving us:
[tex]k=4[/tex]
So, the value of k is 4.
And we're done!
