Solve sqrt{x+4}=3sqrt{x}

$$\sqrt { x + 4 } = 3 \sqrt { x }$$

Solve for x

$$x=\frac{1}{2}=0.5$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{x+4}\right)^{2}=\left(3\sqrt{x}\right)^{2}$$

Calculate \sqrt{x+4} to the power of 2 and get x+4.

$$x+4=\left(3\sqrt{x}\right)^{2}$$

Expand \left(3\sqrt{x}\right)^{2}.

$$x+4=3^{2}\left(\sqrt{x}\right)^{2}$$

Calculate 3 to the power of 2 and get 9.

$$x+4=9\left(\sqrt{x}\right)^{2}$$

Calculate \sqrt{x} to the power of 2 and get x.

$$x+4=9x$$

Subtract 9x from both sides.

$$x+4-9x=0$$

Combine x and -9x to get -8x.

$$-8x+4=0$$

Subtract 4 from both sides. Anything subtracted from zero gives its negation.

$$-8x=-4$$

Divide both sides by -8.

$$x=\frac{-4}{-8}$$

Reduce the fraction \frac{-4}{-8} to lowest terms by extracting and canceling out -4.

$$x=\frac{1}{2}$$

Substitute \frac{1}{2} for x in the equation \sqrt{x+4}=3\sqrt{x}.

$$\sqrt{\frac{1}{2}+4}=3\sqrt{\frac{1}{2}}$$

Simplify. The value x=\frac{1}{2} satisfies the equation.

$$\frac{3}{2}\times 2^{\frac{1}{2}}=\frac{3}{2}\times 2^{\frac{1}{2}}$$

Equation \sqrt{x+4}=3\sqrt{x} has a unique solution.

$$x=\frac{1}{2}$$