I have been struggling on this equation at school, and I seem to find no possible way to do this. Can anyone provide a simple way to find the rules of this equation?
$$x + 3\sqrt{x} - 10 = 0$$
I have been struggling on this equation at school, and I seem to find no possible way to do this. Can anyone provide a simple way to find the rules of this equation?
$$x + 3\sqrt{x} - 10 = 0$$
On
You have $(3\sqrt x)^2=(10-x)^2$ and $9x=100-20x+x^2$, ...., but take care because one of the apparent solution to the quadratic equation won't solve the original equation.
On
Take the term with $\sqrt x$ to the other side and square the equation. That will get rid of the square root and give you a quadratic. Solve the quadratic. Now check whether the solutions satisfy the original equation because squaring can introduce spurious solutions.
On
$$ x-10=-3\sqrt{x} $$
Or $$ x^2+100-20x=9x $$
Or $$x^2 - 29x +100 =0$$
$\text{Apply Quadratic Formula, discard negative 'x' and Njoy!}$
On
$x+3\sqrt{x}$ increases and $10$ is a constant, which says the equation $$x+3\sqrt{x}=10$$ has one root maximum.
But $4$ is a root, which gives the answer.
I will provide one method for you, and after reading the first few steps, you should be able to solve this on your own.
Method: Put the square root on its own side.
$$x - 10 = -3\sqrt{x}$$
Now, squaring both sides, you get
$$(x-10)^2 = 9x$$
You should be able to solve it from there.