Let A2 + 2a-1 = 0, b4-2b2-1 = 0, and 1-ab2 ≠ 0 According to the condition of 1-ab2 ≠ 0, B2 = - A is obtained How did this step come about? Why B & # 178; = - a?

Let A2 + 2a-1 = 0, b4-2b2-1 = 0, and 1-ab2 ≠ 0 According to the condition of 1-ab2 ≠ 0, B2 = - A is obtained How did this step come about? Why B & # 178; = - a?

1.a2+2a-1=0 a2+2a+1-1-1=0 (a+1)2-2=0 (a+1)2=2
2． b4 -2b2 -1=0 b4-2b2 +1-1-1=0 (b2-1)2-2=0 (b2-1)2=2
3. If a + 1 = + / - 2 is obtained from 1 square root, then B2 = 1 + (+ / - 2) is obtained from 2 square root
√ 2 (because 1 - √ 2 ＜ 0, does not hold)
4. From 3, if a + 1 = √ 2, then a = √ 2-1, a B2 = (√ 2-1) (1 + √ 2) = 1
5. In the question, 1-ab2 ≠ 0, and according to 4, we can get 1-ab2 = 1-1 = 0, so we deduce a + 1 = - √ 2, that is, a = - 1 - √ 2 = - (1 + √ 2) = - B2, from above we get B & # 178; = - A
I hope I can understand it. The method may be a little stupid