# find $y$ if $\sqrt{y} + \sqrt{16y} - 2 = 4$

My attempt.

$$\sqrt{16y} = 2y$$

so

$$\sqrt{y}+2y = 6$$ after moving the constants.

I'm not too sure about this, but i think i can get rid of the radical by raising both sides of the equation to the 4th power?

$$y+16y = 1,296$$

After simplification I'm left with this.

$$y = \frac{1,296}{17}$$

• $$\sqrt{16y} = 2\sqrty$$ – J. W. Tanner Nov 17 '19 at 21:10
• It is not true that $(16y)^{\frac{1}{4}}=2y$. It should be $2y^{\frac{1}{4}}$. – IamWill Nov 17 '19 at 21:10
• If you raise both sides of the eqation $\sqrty+2y=6$ to the fourth power you obtain $y+8y\sqrt{y^3}+24y^2\sqrt{y^2}+32y^3\sqrty+16y^4=1296$. Then again, identity $\sqrt{16y}=2y$ is not a thing, so whatever. – Gae. S. Nov 17 '19 at 21:14

Recognize $$\sqrt{16}=2$$ and simplify the equation
$$3\sqrt{y}= 6$$
$$y = 16$$
Let $$x = \sqrt{y}$$. Then your equation is equivalent to $$x+2x-2=4$$ Thus: $$3x = 6 \Rightarrow x = 2.$$ Now, we know $$x = 2 = \sqrt{y}$$,so it follows $$y = 16$$.