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(D) Given that the equation $x^2+2ax+b^2=0$ has two distinct real roots,its discriminant $D_1 > 0$.$D_1 = (2a)^2 - 4(1)(b^2) = 4a^2 - 4b^2 > 0 \Righta

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no real roots

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(D) Given that the equation $x^2+2ax+b^2=0$ has two distinct real roots,its discriminant $D_1 > 0$.$D_1 = (2a)^2 - 4(1)(b^2) = 4a^2 - 4b^2 > 0 \Rightarrow a^2 > b^2$  $(i)$Similarly,for the equation $x^2+2bx+c^2=0$,the discriminant $D_2 > 0$.$D_2 = (2b)^2 - 4(1)(c^2) = 4b^2 - 4c^2 > 0 \Rightarrow b^2 > c^2$  $(ii)$From $(i)$ and $(ii)$,we have $a^2 > b^2 > c^2$,which implies $a^2 > c^2$.Now,consider the equation $x^2+2cx+a^2=0$. Its discriminant $D_3$ is:$D_3 = (2c)^2 - 4(1)(a^2) = 4c^2 - 4a^2 = 4(c^2 - a^2)$.Since $a^2 > c^2$,it follows that $c^2 - a^2 Therefore,the equation $x^2+2cx+a^2=0$ has no real roots.

Suppose $a, b, c$ are real numbers,and each of the equations $x^2+2ax+b^2=0$ and $x^2+2bx+c^2=0$ has two distinct real roots. Then,the equation $x^2+2cx+a^2=0$ has

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