Solving a quadratic equation with square root constants.

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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?

$$x + 3\sqrt{x} - 10 = 0$$

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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.

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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.

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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.

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$$ x-10=-3\sqrt{x} $$

Or $$ x^2+100-20x=9x $$

Or $$x^2 - 29x +100 =0$$

$\text{Apply Quadratic Formula, discard negative 'x' and Njoy!}$

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$x+3\sqrt{x}$ 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.

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Further to other suggestions, write $x=y^2$ so that $$y^2+3y-10=0$$and solve as a normal quadratic.

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Remember that when ever there is a square root on $x$ not $x^2$ you are supposed to take it to righthand side of the equation and then you have to square the whole equation. In this way the root would be removed from the equation and you will be left with a solvable equation.