Given the solution of the quadratic equation AX ^ 2 + BX + C = 0 about x? You have already answered the answer to this question, but can you tell us the detailed process, thank you. We know that the two univariate quadratic equations ax ^ 2 + BX + C = 0 about X are α and β, and the two equations x ^ 2 + (α + 1) x + β ^ 2 = 0 and x ^ 2 + (β + 1) x + α ^ 2 = 0 about X have a unique common root, so we can find the relationship of a, B and C The answer might be - B / a = - 2 / 5, C / a = - 3 / 25

Given the solution of the quadratic equation AX ^ 2 + BX + C = 0 about x? You have already answered the answer to this question, but can you tell us the detailed process, thank you. We know that the two univariate quadratic equations ax ^ 2 + BX + C = 0 about X are α and β, and the two equations x ^ 2 + (α + 1) x + β ^ 2 = 0 and x ^ 2 + (β + 1) x + α ^ 2 = 0 about X have a unique common root, so we can find the relationship of a, B and C The answer might be - B / a = - 2 / 5, C / a = - 3 / 25

You mean: the equations x ^ 2 + (α + 1) x + β ^ 2 = 0 and x ^ 2 + (β + 1) x + α ^ 2 = 0
They all have one root and are equal, right?
If each equation has more than one root and one common root, it is not like this
α. β is the two unequal real roots of the equation AX ^ 2 + BX + C = 0
So: α + β = - B / A, α β = C / A
And: (α + 1) ^ 2-4 β ^ 2 = 0 --- (1)
(β+1)^2-4α^2=0--------(2)
(1)-(2)：(α+1+β+1)(α+1-β-1)=4(β+α)(β-α)
So: α + β + 2 = - 4 (β + α)
That is: α + β = - 2 / 5
(1)+(2)：α^2+β^2+2(α+β)+2=4(α^2+β^2)
That is: 3 (α ^ 2 + β ^ 2) = 2 (α + β) + 2 = 6 / 5
That is: (α + β) ^ 2-2 α β = 2 / 5
That is: 2 α β = 4 / 25-2 / 5 = - 6 / 25
That is: α β = - 3 / 25
So: B / a = 2 / 5, C / a = - 3 / 25