How solve this type of equations?

Question

Somebody tell me that I can put inside the square root the x, like an $x^2$, but I don't know how to do that

$x\sqrt{1+x}=\frac{12}{15}$

Answer 1

First of all, you can easily prove that $x>0$.

Then, you can use the following rules:

• $x=\sqrt{x^2}$
• $\sqrt a \sqrt b= \sqrt{ab}$

Answer 2

$$x\sqrt{1+x}=\frac{12}{15}=\sqrt{x^2(1+x)}=\frac{12}{15}$$

Note how we squared $x$ when we brought it into the radical.

Next simplify the RHS.

$$\frac{12}{15}=\frac{4}{5}$$

Now put the right hand side in a radical by squaring it.

$$\sqrt{x^2(1+x)}=\sqrt{(\frac{4}{5})^2}$$

Now, you can simply remove the radicals.

$$x^2(1+x)=(\frac{4}{5})^2$$

With the radicals gone, can you expand each side and solve for $x$?

Answer 3

HINT:

$$x\sqrt{1+x}=\frac{12}{15} \Longleftrightarrow$$ $$x\sqrt{1+x}=\frac{4}{5} \Longleftrightarrow$$ $$\sqrt{1+x}=\frac{4}{5x} \Longleftrightarrow$$ $$1+x=\left(\frac{4}{5x}\right)^2 \Longleftrightarrow$$ $$1+x=\frac{16}{25x^2} \Longleftrightarrow$$ $$25x^2(x+1)=16 \Longleftrightarrow$$ $$x^3+x^2=\frac{16}{25} \Longleftrightarrow$$ $$\frac{1}{25}\left(25x^3+25x^2-16\right)=0 \Longleftrightarrow$$ $$25x^3+25x^2-16=0 \Longleftrightarrow$$