It is proved that the value of X & # 178; - 4x27 is greater than zero by matching method Solving x square + 2ax-b square + a square = 0 by collocation method If a, B satisfy a ^ B ^ + A ^ + B ^ + 10ab + 16 = 0 Finding the value of a and B

It is proved that the value of X & # 178; - 4x27 is greater than zero by matching method Solving x square + 2ax-b square + a square = 0 by collocation method If a, B satisfy a ^ B ^ + A ^ + B ^ + 10ab + 16 = 0 Finding the value of a and B

x²-4x+27
=x²-4x+4+23
=(x-2)²+23
The square term is constant and nonnegative, (X-2) &# 178; ≥ 0 (X-2) &# 178; + 23 ≥ 23 > 0
The value of X & # 178; - 4x + 27 is always greater than zero
x²+2ax-b²+a²=0
(x+a)²=b²
X + a = B or x + a = - B
X = B-A or x = - B-A
a²b²+a²+b²+10ab+16=0
(a²b²+8ab+16)+(a²+2ab+b²)=0
(ab+4)²+(a+b)²=0
The sum of the two nonnegative terms is 0, and the two nonnegative terms are 0
ab+4=0 a+b=0
The solution is a = 2, B = - 2 or a = - 2, B = 2