How do you solve (2q)/(2q+3)-(2q)/(2q-3)=12q2q+3−2q2q−3=1?
1 Answer
We should find a common denominator of
(2q(2q-3))/((2q+3)(2q-3))-(2q(2q+3))/((2q+3)(2q-3))=((2q+3)(2q-3))/((2q+3)(2q-3))2q(2q−3)(2q+3)(2q−3)−2q(2q+3)(2q+3)(2q−3)=(2q+3)(2q−3)(2q+3)(2q−3)
Combining:
(2q(2q-3)-2q(2q+3))/(4q^2-9)=(4q^2-9)/(4q^2-9)2q(2q−3)−2q(2q+3)4q2−9=4q2−94q2−9
Multiplying through and disregarding the denominator since they're all equal:
(4q^2-6q)-(4q^2+6q)=4q^2-9(4q2−6q)−(4q2+6q)=4q2−9
Pay attention to positives and negatives here:
-12q=4q^2-9−12q=4q2−9
4q^2+12q-9=04q2+12q−9=0
Using the Quadratic Formula:
q=(-b+-sqrt(b^2-4ac))/(2a)=(-12+-sqrt(144+144))/8q=−b±√b2−4ac2a=−12±√144+1448
q=(-12+-12sqrt2)/8=(-3+-3sqrt2)/4q=−12±12√28=−3±3√24