Question

Use the properties of proportion; Solve for y.

√y+1+√y-1 /√y+1-√y-1= 4y-1/2

Answer 1

Answer:

$\frac{\sqrt{y+1+\sqrt{y}-1}}{\sqrt{y+1-\sqrt{y}-1}}=\frac{4y-1}{2}\quad :\quad y\approx \:1.65913\dots$

Step-by-step explanation:

$\mathrm{\frac{\sqrt{y+1+\sqrt{y}-1}}{\sqrt{y+1-\sqrt{y}-1}}=\frac{4y-1}{2}}$

$\mathrm{Apply\:fraction\:cross\:multiply:\:if\:}\frac{a}{b}=\frac{c}{d}\mathrm{\:then\:}a\cdot \:d=b\cdot \:c$

$\sqrt{y+1+\sqrt{y}-1}\cdot \:2=\sqrt{y+1-\sqrt{y}-1}\left(4y-1\right)$

$\sqrt{y+1+\sqrt{y}-1}\cdot \:2=4\sqrt{y+1-\sqrt{y}-1}y-\sqrt{y+1-\sqrt{y}-1}$

$4\left(y+\sqrt{y}\right)=\left(-\sqrt{y}+y\right)\left(4y-1\right)^2$

$4y+4\sqrt{y}=-16y^2\sqrt{y}+8y\sqrt{y}-\sqrt{y}+16y^3-8y^2+y$

$\mathrm{Subtract\:}4y\mathrm{\:from\:both\:sides}$

$4y+4\sqrt{y}-4y=-16y^2\sqrt{y}+8y\sqrt{y}-\sqrt{y}+16y^3-8y^2+y-4y$

$4\sqrt{y}=-16y^2\sqrt{y}+8y\sqrt{y}-\sqrt{y}+16y^3-8y^2-3y$

$\mathrm{Subtract\:}-16y^2\sqrt{y}+8y\sqrt{y}-\sqrt{y}\mathrm{\:from\:both\:sides}$

$4\sqrt{y}-\left(-16y^2\sqrt{y}+8y\sqrt{y}-\sqrt{y}\right)=-16y^2\sqrt{y}+8y\sqrt{y}-\sqrt{y}+16y^3-8y^2-3y-\left(-16y^2\sqrt{y}+8y\sqrt{y}-\sqrt{y}\right)+8y\sqrt{y} -\sqrt{y}$

$16y^2\sqrt{y}-8y\sqrt{y}+5\sqrt{y}=16y^3-8y^2-3y$

$\sqrt{y}\left(16y^2-8y+5\right)=16y^3-8y^2-3y$

$\mathrm{Divide\:both\:sides\:by\:}16y^2-8y+5$

$\frac{\sqrt{y}\left(16y^2-8y+5\right)}{16y^2-8y+5}=\frac{16y^3}{16y^2-8y+5}-\frac{8y^2}{16y^2-8y+5}-\frac{3y}{16y^2-8y+5}$

$\sqrt{y}=\frac{16y^3-8y^2-3y}{16y^2-8y+5}$

$\mathrm{y=\frac{256y^6-256y^5-32y^4+48y^3+9y^2}{256y^4-256y^3+224y^2-80y+25}}$

$y=0,\:y\approx \:1.65913\dots$