If the equation a(b-c)x^2 + b(c-a)x + c(a-b) | vedclass

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(A) Given the quadratic equation $a(b-c)x^2 + b(c-a)x + c(a-b) = 0$. Notice that the sum of the coefficients is $a(b-c) + b(c-a) + c(a-b) = ab - ac +

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$117$

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(A) Given the quadratic equation $a(b-c)x^2 + b(c-a)x + c(a-b) = 0$. Notice that the sum of the coefficients is $a(b-c) + b(c-a) + c(a-b) = ab - ac + bc - ab + ac - bc = 0$. Since the sum of coefficients is $0$,$x = 1$ is a root of the equation. Because the roots are equal,both roots must be $1$. The product of the roots is $\frac{c(a-b)}{a(b-c)} = 1 	imes 1 = 1$. Thus,$c(a-b) = a(b-c)$ $\Rightarrow ac - bc = ab - ac$ $\Rightarrow 2ac = ab + bc = b(a+c)$. Given $a+c = 15$ and $b = \frac{36}{5}$,we have $2ac = \frac{36}{5} 	imes 15 = 36 	imes 3 = 108$. So,$ac = 54$. We need to find $a^2 + c^2$. Using the identity $a^2 + c^2 = (a+c)^2 - 2ac$,we get $a^2 + c^2 = (15)^2 - 108 = 225 - 108 = 117$.

If the equation $a(b-c)x^2 + b(c-a)x + c(a-b) = 0$ has equal roots,where $a + c = 15$ and $b = \frac{36}{5}$,then $a^2 + c^2$ is equal to . . . . . .

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