How do you solve #(x ^ { 2} + 3) ^ { 2} + 21= 10x ^ { 2} + 30#?
1 Answer
Dec 10, 2016
At first glance, this problem looks very difficult. However, you just have to expand and see what happens...
#(x^2 + 3)(x^2 + 3) + 21 - 10x^2 - 30 = 0#
#x^4 + 6x^2 + 9 + 21 - 10x^2 - 30 = 0#
#x^4 - 4x^2 + 9 -9 = 0#
#x^4 - 4x^2 = 0#
#x^2(x^2 - 4) = 0#
#x = 0 and +-2#
Hopefully this helps!
