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(B) The given equation is $(a^2+b^2)x^2-2b(a+c)x+(b^2+c^2)=0$.This can be rewritten as $a^2x^2-2abx+b^2+b^2x^2-2bcx+c^2=0$.This simplifies to $(ax-b)^

Vedclass · Accepted Answer

$36$

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(B) The given equation is $(a^2+b^2)x^2-2b(a+c)x+(b^2+c^2)=0$.This can be rewritten as $a^2x^2-2abx+b^2+b^2x^2-2bcx+c^2=0$.This simplifies to $(ax-b)^2+(bx-c)^2=0$.Since $a, b, c$ are real,we must have $ax-b=0$ and $bx-c=0$,which implies $x = b/a = c/b$.Thus,$b^2 = ac$,meaning $a, b, c$ are in geometric progression.For $a, b, c$ to be sides of a triangle,they must satisfy the triangle inequality: $a+b > c$,$b+c > a$,and $c+a > b$.Substituting $b = ax$ and $c = bx = ax^2$:$a + ax > ax^2$ $\Rightarrow x^2 - x - 1 $ax + ax^2 > a$ $\Rightarrow x^2 + x - 1 > 0$ $\Rightarrow x > \frac{-1+\sqrt{5}}{2}$ (since $x > 0$).$a + ax^2 > ax \Rightarrow x^2 - x + 1 > 0$ (always true).Combining these,we get $\frac{\sqrt{5}-1}{2} So,$\alpha = \frac{\sqrt{5}-1}{2}$ and $\beta = \frac{\sqrt{5}+1}{2}$.Then $\alpha^2 + \beta^2 = \frac{5+1-2\sqrt{5}}{4} + \frac{5+1+2\sqrt{5}}{4} = \frac{12}{4} = 3$.Therefore,$12(\alpha^2+\beta^2) = 12(3) = 36$.

Let $a, b, c$ be the lengths of three sides of a triangle satisfying the condition $(a^2+b^2)x^2-2b(a+c)x+(b^2+c^2)=0$. If the set of all possible values of $x$ is the interval $(\alpha, \beta)$,then $12(\alpha^2+\beta^2)$ is equal to.

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