Solve for the indicated variable. Assume the variable you are solving for must be greater than 0.
-A $$A = w \left( \frac { p - 2 w } { 2 } \right) , \text { for } w$$
A) $$w = \frac { p + \sqrt { p ^ { 2 } - 16 A } } { 4 }$$
B) $$w = \frac { 2 p + \sqrt { p ^ { 2 } - 16 A } } { 4 }$$
C) $$w = \frac { - 2 p + \sqrt { p ^ { 2 } + 16 A } } { 4 }$$
D) $$w = \frac { p + ( p - 4 \sqrt { \mathrm { A } } ) } { 2 }$$

Question

Quizplus · Accepted Answer

To solve for $w$, we start with the given equation $A = w \left( \frac { p - 2 w } { 2 } \right)$. Multiplying both sides by 2 to eliminate the denominator gives $2A = w(p - 2w)$. Expanding and rearranging terms to form a quadratic equation in terms of $w$, we get $2w^2 - pw + 2A = 0$. Using the quadratic formula, $w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 2$, $b = -p$, and $c = 2A$, we find $w = \frac{p \pm \sqrt{p^2 - 4(2)(2A)}}{4}$, which simplifies to $w = \frac{p \pm \sqrt{p^2 - 16A}}{4}$. Since $w > 0$, we choose the positive root, leading to the correct answer: $w = \frac{p + \sqrt{p^2 - 16A}}{4}$.

Solve for the Indicated Variable A=w(p−2w2), for wA = w \left( \frac { p - 2 w } { 2 } \right) , \text { for } wA=w(2p−2w​), for w

Solve for the Indicated Variable $A = w \left( \frac { p - 2 w } { 2 } \right) , \text { for } w$