How to solve 15 & # 178; + X & # 178; = (25-x) &# 178; + 10 & # 178

How to solve 15 & # 178; + X & # 178; = (25-x) &# 178; + 10 & # 178

225+X^2=625-50X+X^2+100
50X=625-225+100
50X=500
X=10.

X & # 178; = 15, find the value of X

x²=15
x=±√15

If √ 35-x & # 178; - √ 15-x & # 178; = 4, find the value of √ 35-x & # 178; + √ 15-x & # 178

√(35-x²)-√(15-x²)=[√(35-x²)-√(15-x²)][√(35-x²)+√(15-x²)]/[√(35-x²)+√(15-x²)]
=[(35-x²)-(15-x²)]/[√(35-x²)+√(15-x²)]
=20/[√(35-x²)+√(15-x²)]=4
√(35-x²)+√(15-x²)=20/4=5
Hint: it is very simple, that is, the molecule is rationalized, and the denominator is √ (35-x & # 178;) + √ (15-x & # 178;)