Help! I tried plugging in numbers.... pls help

🙏🙏🙏

-\(tommarvoloriddle\)

tommarvoloriddle Jul 7, 2019

#1**+7 **

NVM! I solved this question by solving for x, thank you for helping, melody!

tommarvoloriddle Jul 7, 2019

#2**+2 **

Here is the graph

https://www.desmos.com/calculator/7r6zlevfrv

Using the graph can you see what the answer is?

You also need to be able to work these out without graphing first (it is good to graph afterwards for checking you answer)

X^2 is always going to be positive.

x can't be 1

Do you know how to use calculus Tom?

If so it is easy just to differentiate and then set y'=0 to find the stationary points etc.

Melody Jul 7, 2019

#5**+2 **

It doesn't actually matter what the solution to this quadratic is. The only thing that is important is that we find all values such that this quadratic (in ) has a real solution. For that we look at the discriminant.

We see that we can only find a suitable value of x if \( y^2 + 4y \ge 0\)

tommarvoloriddle
Jul 7, 2019

#9**+2 **

ok, so i figured that you didn't need to solve for x... whoops... and you could use the dicriminant...

tommarvoloriddle
Jul 7, 2019