How do you solve #w ^ { 2} + 4m = 45#?

1 Answer
Dec 10, 2017

#w = +- sqrt(45 - 4m)#
#m = (45- w^2) // 4#

Explanation:

With a some rearranging you can express #w# in function of #m# and #m# in function of #w#

Solve for #w# :

#w^2 + 4m = 45#

Subtract #4m# to both sides
#w^2 + cancel(4m - 4m) = 45 - 4m#

Take the square root of both sides
#sqrt(w^2) = +- sqrt(45 - 4m)#

We know that #AA x; sqrt(x^2) = x#
#w = +- sqrt(45 - 4m)#

Solve for #m# :

#w^2 + 4m = 45#

Subtract #w^2# to both sides
#cancel(w^2 - w^2) + 4m = 45 - w^2#

Divide both sides by #4#
#cancel(4)m // cancel(4) = (45 - w^2) // 4#

Write correctly the equation
#m = (45- w^2) // 4#