If the difference between the two equations 2x ^ 2 - (k-1) x + K + 3 = 0 is 1, what is the value of K? The absolute value of (x1-x2) = the absolute value of B ^ 2-4ac / a [this is a theorem. So, 1 = (k-1) ^ 2-8 (K + 3) It's wrong. I must be very wrong! I'd like to know what's wrong with me.

If the difference between the two equations 2x ^ 2 - (k-1) x + K + 3 = 0 is 1, what is the value of K? The absolute value of (x1-x2) = the absolute value of B ^ 2-4ac / a [this is a theorem. So, 1 = (k-1) ^ 2-8 (K + 3) It's wrong. I must be very wrong! I'd like to know what's wrong with me.

So, 1 = (k-1) ^ 2-8 (K + 3), should be 4 = (k-1) ^ 2-8 (K + 3) two for x1, x2x1 + x2 = (k-1) / 2x1 * x2 = (K + 3) / 2x1-x2 = 1 (x1-x2) ^ 2 = 1 (x1 + x2) ^ 2-4x1x2 = 1 (k-1) ^ 2 / 4-2 (K + 3) = 1 (k-1) ^ 2-8 (K + 3) - 4 = 0k ^ 2-10k-27 = 0, then do it yourself