If the roots of the equation $\left ({a}^{2}+{b}^{2}\right){x}^{2}+2\left (bc + ad\right)x+\left ({c}^{2}+{d}^{2}\right)=0$ are real then ${a}^{2},bd,{c}^{2}$ are in
- $AP$
- $GP$
- $HP$
- $AGP$
For the quadratic equation (a^2+b^2)x^2 + 2(bc+ad)x + (c^2+d^2) = 0 to have real roots, the discriminant D must be >= 0. D = [2(bc+ad)]^2 - 4(a^2+b^2)(c^2+d^2) = 4(b^2c^2 + a^2d^2 + 2abcd) - 4(a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2) = 4(-a^2c^2 + 2abcd - b^2d^2) = -4(ac-bd)^2. For D >= 0, we must have ac=bd, which implies a/b = d/c, or ad=bc. This condition implies a^2, bd, c^2 are in GP if a^2/bd = bd/c^2, which is a^2c^2 = b^2d^2, consistent with ac=bd.
For the roots of the quadratic equation (a^2 + b^2)x^2 + 2(bc + ad)x + (c^2 + d^2) = 0 to be real, the discriminant must be greater than or equal to zero. Setting the discriminant D = (2(bc + ad))^2 - 4(a^2 + b^2)(c^2 + d^2) gives 4(b^2c^2 + 2abcd + a^2d^2) - 4(a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2). Dividing by 4 and simplifying leaves the inequality 2abcd - a^2c^2 - b^2d^2 greater than or equal to 0, which can be rewritten by factoring out a negative sign as (ad - bc)^2 less than or equal to 0. Because the square of any real number cannot be negative, the expression (ad - bc)^2 must exactly equal zero, meaning ad = bc. Dividing both sides by cd yields the ratio a/c = b/d, which we can call k, allowing us to express a as ck and b as dk. Substituting these into the sequence a^2, bd, c^2 gives (ck)^2, d(dk), c^2, which simplifies to c^2k^2, dk^2, c^2. This shows that multiplying the middle term by c^2 equals the product of the first and third terms, since c^2 times dk^2 equals c^2k^2 times d, proving the terms are in geometric progression.