a2 + b2 cannot be factorised in real numbers, only in complex numbers. The fact that you ask this question indicates that you are nowhere near old enough to be studying complex numbers.
a2 + b2 cannot be factorised in real numbers, only in complex numbers. The fact that you ask this question indicates that you are nowhere near old enough to be studying complex numbers.
a2 + b2 cannot be factorised in real numbers, only in complex numbers. The fact that you ask this question indicates that you are nowhere near old enough to be studying complex numbers.
a2 + b2 cannot be factorised in real numbers, only in complex numbers. The fact that you ask this question indicates that you are nowhere near old enough to be studying complex numbers.
0
a2 - 4a + 4
l a2 b2 is c2!!Its completely norma
A2 + B2 = C2 If C=8, then A2 + B2 = 64
(a3 + b3)/(a + b) = (a + b)*(a2 - ab + b2)/(a + b) = (a2 - ab + b2)
a2+2a2b+2ab2+b2
0
a2
a2 - 4a + 4
No. If you expand (a + b)2 you get a2 + 2ab + b2. This is not equal to a2 + b2
l a2 b2 is c2!!Its completely norma
sqrt(a2 + b2) can't be simplified. Neither can (a2 + b2) .
The reciprocal of a + bi is a - bi:1/(a + bi) since the conjugate is a - bi:= 1(a - bi)/[(a + bi)(a - bi)]= (a - bi)/[a2 - (b2)(i2)] since i2 equals to -1:= (a - bi)/(a2 + b2) since a2 + b2 = 1:= a - bi/1= a - bi
A2 + B2 = C2 If C=8, then A2 + B2 = 64
(a3 + b3)/(a + b) = (a + b)*(a2 - ab + b2)/(a + b) = (a2 - ab + b2)
a2+b2+c2=x2+y2+z2 divide each side by 2 (a2+b2+c2)/2=(x2+y2+z2)/2 a+b+c=x+y+z
To evaluate it you need to know both b and a. But it is possible to factorise it as follows: b2-a2 = (b-a)*(b+a)