XIII OM - I - Problem 6

Factor the quadratic polynomial into real factors






where $ p $ and $ q $ are real numbers satisfying the inequality











Please note that the mathematical expressions and symbols are kept as they are, only the text has been translated.
From the inequality $ p^2 - 4q < 0 $, it follows that the quadratic polynomial $ y^2 + py + q $ has no real roots; therefore, the quartic polynomial $ x^4 + px^2 + q $ also has no real roots. If the number $ x $ were a real root of the second polynomial, then the number $ y = x^2 $ would be a real root of the first polynomial. Therefore, the desired factorization can only have the form

Comparing the coefficients of the same powers of $ x $ on both sides of the above equality, we obtain the equations

\[ (1) \qquad  \alpha + \gamma = 0, \]

\[ (2) \qquad  \beta + \delta + \alpha \gamma = p, \]

\[ (3) \qquad  \alpha \delta + \beta \gamma = 0, \]

\[ (4) \qquad  \beta \delta = q. \]

From equation (1) we have $ \gamma = - \alpha $, substituting into equation (3) gives

a) If $ \alpha = 0 $, then from equations (2) and (4) it follows that

This system of equations has no real solutions, because $ p^2 - 4q < 0 $.
b) $ \delta = \beta $; from equation (4) we then obtain

The above values of $ \beta $ and $ \delta $ are real, since from the inequality $ p^2 - 4q < 0 $ it follows that $ 4q > p^2 $, hence $ q > 0 $.
According to equation (2) $ \alpha \gamma = p - \beta - \delta $, so to determine $ \alpha $ and $ \gamma $ we have the system of equations

Therefore

From the inequality $ 4q > p^2 $ it follows that $ 2 \sqrt{q} > |p| \geq p $, hence $ 2 \sqrt{q} - p > 0 $; Since $ 4q - p^2 = (2 \sqrt{q} - p) (2 \sqrt{q} + p) > 0 $, then also $ 2 \sqrt{q} + p > 0 $.
To obtain a real value for $ \alpha $, we need to take the upper sign in the above formula for $ \alpha^2 $. We obtain the following real values of the coefficients $ \alpha $, $ \beta $, $ \gamma $, $ \delta $:

The desired factorization is expressed by the formula: