Question

$$\frac { - 2 x } { \sqrt { 4 - x ^ { 2 } } } \cdot \frac { \sqrt { 4 } } { \sqrt { 4 - x ^ { 2 } } = v }$$

Answer

x=sqrt(-v*d+16*Te*t),-sqrt(-v*d+16*Te*t)

Solution


Simplify  \(4Tet\times 4\)  to  \(16tTe\).
\[16tTe-{x}^{2}=vd\]
Regroup terms.
\[16Tet-{x}^{2}=vd\]
Subtract \(16Tet\) from both sides.
\[-{x}^{2}=vd-16Tet\]
Multiply both sides by \(-1\).
\[{x}^{2}=-vd+16Tet\]
Take the square root of both sides.
\[x=\pm \sqrt{-vd+16Tet}\]