If the difference between the absolute values of the two equations a + B = a + x 2 + B is equal

If the difference between the absolute values of the two equations a + B = a + x 2 + B is equal

Let two of the equations x2 + ax + B = 0 be x1, X2; two of the equations x2 + BX + a = 0 be X3, x4, then X1 + x2 = - A, X1 * x2 = B; X3 + X4 = - B, X3 * X4 = AX2 + ax + B = 0 | x1-x2 | = √ (x1-x2) ^ 2 = √ (x1 + x2) ^ 2-4x1 * x2 = √ a ^ 2-4bx2 + BX + a = 0 | x3-x4 | =

Let the equation | x2 + ax | = 4 have only three unequal roots of real numbers. Find the value of a and the corresponding three roots

∵| - x2 + ax | = 4, ∵ x2 + ax-4 = 0 (1) or x2 + ax + 4 = 0 (2), equation (1) and (2) can not have the same root, but the original equation has three unequal real roots, ∵ one of equation (1) and (2) has equal root, and △ 1 = A2 + 16 ＞ 0, ∵ 2 = a2-16 = 0, ∵ a = ± 4. When a = 4, the original equation is x2 + 4x-4 = 0 or x2 + 4x + 4 = 0, the solution of the original equation is x = - 2, - 2 ± 22; when a = - 4, the original equation is x2-4x-4 = 0 or x2-4x+ 4 = 0, the solution of the original equation is: x = 2,2 ± 22;

And at least one of the absolute values of two X1 and X2 of equation x2 + ax + B = 0 is not less than 1, and | a | + | B | ≤ 1, it is proved that | a | + | B | = 1

We only need to prove that | a | + | B | > = 1
We know | a | + | B | = | X1 + x2 | + | X1 * x2 from Weida's theorem (relation between root and coefficient) X1 + x2 = - A, X1 * x2 = B|
=|x1+x2|+|x1|*|x2|
Let | x1 | > = 1, then the above formula > = | X1 + x2 | + | x2 | > = | X1 + x2-x2 | = | x1 | > = 1
The proof is over