Question

Solve the exercise

$16y^2+12=-32x$

Short answer

There is no solution to the equation.

Step-by-step solution

Step 1. Given information

The given equation$16y^2+12=-32x$

Step 2. Simplify the quadratic equation  

Simplifying the equation$$\begin{gathered} (a-b)(a+b)=0 \\ a-b=0ora+b=0 \\ a=ba=bora=-b \\ \mathrm{Simplify}\mathrm{the}\mathrm{quadratic}\mathrm{expression}: \\ 164y^2+12=-32y[\mathrm{Given}\mathrm{the}\mathrm{expression}] \\ 16{\mathrm{y}}^2+32\mathrm{y}+12=0[\mathrm{Write}\mathrm{the}\mathrm{quadratic}\mathrm{equation}\mathrm{of}\mathrm{standard}\mathrm{form}] \\ 4(4{\mathrm{y}}^2+8\mathrm{y}+3)=0[\mathrm{Factor}\mathrm{out}4\mathrm{from}\mathrm{expression}] \\ 4(4{\mathrm{y}}^2+6\mathrm{y}+2\mathrm{y}+3)=0[\mathrm{Since}8\mathrm{y}=6\mathrm{y}+2\mathrm{y}] \\ 4(2\mathrm{y}(2\mathrm{y}+3)+1(2\mathrm{y}+3\left)\right)=0[\mathrm{Factor}\mathrm{out}\mathrm{the}\mathrm{common}\mathrm{factor}\mathrm{from}\mathrm{group}] \\ 4(2\mathrm{y}+1)(2\mathrm{y}+3)=0[\mathrm{Factor}\mathrm{the}\mathrm{common}\mathrm{factor}\mathrm{from}\mathrm{expression}] \\ 4\simeq 0\mathrm{or}2\mathrm{y}+1=0\mathrm{or}2\mathrm{y}+3=0[\mathrm{Use}\mathrm{the}\mathrm{zero}\mathrm{product}\mathrm{property}] \\ ory=-\frac{1}{2}ory=-\frac{3}{2} \end{gathered}$$

Step 3. Checking the equation

$$\begin{gathered} \mathrm{Check} \\ \mathrm{Substitute}\mathrm{the}\mathrm{value}\mathrm{y}=-\frac{1}{2}\mathrm{In}\mathrm{the}\mathrm{original}\mathrm{equation}. \\ 16{\mathrm{y}}^2+12=-32[\mathrm{Given}\mathrm{equation}] \\ 16(-\frac{1}{2}{)}^2+12=-32\frac{-1}{2}*[\mathrm{Put}\mathrm{y}=-1/2] \\ 4+16=16 \\ 20=16\mathrm{False} \\ \mathrm{Substitute}\mathrm{the}\mathrm{value}\mathrm{y}=-\frac{3}{2}\mathrm{in}\mathrm{the}\mathrm{original}\mathrm{equation}. \\ 16(-\frac{3}{2}{)}^2+12=-32\frac{-3}{2}[\mathrm{Put}\mathrm{y}=-3/2] \\ 48+12=48 \\ 60=16\mathrm{False} \\ \mathrm{Hence}\mathrm{there}\mathrm{is}\mathrm{no}\mathrm{solution} \end{gathered}$$