+296 \(\sqrt{53+20\sqrt{7}}\) can be written in the form \(a+b\sqrt{c}\), where \(a, b\) and \(c\) are integers and \(c\) has no factors which is a perfect square of any positive integer other than 1. Find \(a+b+c\).
+128408 √ [53 + 20√ 7] = a + b√ c square both sides
53 + 20√ 7 = a^2 +2ab√ c + b^2c
Equating terms
2ab = 20 a^2 + b^2c = 53
ab =10 a^2 + 7b^2 =53
And c = 7
Then
a^2 + 7b^2 = 53
Let b = 2
a^2 + 7(2)^2 = 53
a^2 + 28 = 53
a^2 = 25
a = 5
So
a + b√ c = 5 + 2√ 7
So
a + b + c = 5 + 2 + 7 = 14
