Because: 5x? -24? Px+5q=0 has two equal real roots,

So there are: b? -4ac=(-24？ p)？ -4x5x5q=0，

D: p？ =25/6q≥0。 Because (p≠0), p? = 25/6q & gt； 0, that is, q>0.

Equation x? +px+q=0 has:

b？ -4ac=p？ -4q = 25/6q-4q = 1/6q & gt； 0.

So equation x? +px+q=0 has two unequal real roots.

The second solution: (reduction to absurdity)

Suppose that at most one of the four equations has two solutions, that is, only one of the four equations has two different solutions or no equation has two solutions. Only one has been disproved, and the following can be obtained in the same way. )

That is, the four equations are: b? -4ac≤0 or 3 4ac≤0 in 4 equations:

There are: (2a+b)-2 √ CD ≤ 0, (2b+c)-2 √ AD ≤ 0, (2c+d)-2 √ AB ≤ 0, (2d+a)-2 √ BC ≤ 0.

Add the two sides and move again: 3a+3b+3c+3d ≤ 2 √ CD+2 √ AD+2 √ AB+2 √ BC. . . . . ①

According to the theorem: m+n ≥ 2 √ Mn (m > 0，n & gt0)

so 3a+3 b+3c+3d =(a+b)+(a+d)+(b+c)+(c+d)+a+b+ c+d >； 2√cd+2√ad+2√ab+2√bc .。 . . ②

① contradicts ②, so the proposition does not hold.

Therefore, at least two of the four equations have unequal real roots.

Thank you. hope this helps