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Question:

If $\;a,\, b,\, c,\, d \;$ are real and $\;ac = 2( b + d) $ ,then show that at least one of the equations: $\; x^2 + ax + b = 0 \;$ and $\; x^2 + cx + d = 0 \;$ has real roots.

I've tried to solve this by adding both the discriminants which gave me $ (a - c)^2$ . Now how to put a logic that show one of the discriminants is real which will give at least real roots of one equation?

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    $\begingroup$ $(a-c)^2\ge 0$; what does that tell you about the signs of the two discriminants? $\endgroup$ – rogerl Jul 2 '14 at 17:58
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$(a - c)^2 \ge 0$, because every square of a real number is nonnegative.

Therefore, $a^2 - 2ac + c^2 \ge 0$, or $a^2 + c^2 - 4b - 4d \ge 0$, so $(a^2 - 4b) + (c^2 - 4d) \ge 0$. As both terms in this sum are real, and their sum is nonnegative, at least one of the terms must be nonnegative, and as these are the discriminants of the two equations, this gives that at least one of the equations has a nonnegative discriminant and therefore real roots.

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First equation's discriminant is

$$a^2-4b$$

Second equation's discriminant is

$$c^2-4d$$

If both discriminants are negative so is their sum, which is

$$a^2-4(b+d)+c^2=a^2-2ac+c^2=(a-c)^2$$

Since any squared real is non negative you get a direct contradiction.

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