The product of all the real roots of x^2-8x+9 | vedclass

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(B) Given equation: $x^2-8x+9-\frac{8}{x}+\frac{1}{x^2}=0$ Rearranging the terms: $(x^2+\frac{1}{x^2}) - 8(x+\frac{1}{x}) + 9 = 0$ Let $t = x+\frac{1}

Vedclass · Accepted Answer

$1$

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(B) Given equation: $x^2-8x+9-\frac{8}{x}+\frac{1}{x^2}=0$ Rearranging the terms: $(x^2+\frac{1}{x^2}) - 8(x+\frac{1}{x}) + 9 = 0$ Let $t = x+\frac{1}{x}$. Then $t^2 = x^2+2+\frac{1}{x^2}$,so $x^2+\frac{1}{x^2} = t^2-2$. Substituting into the equation: $(t^2-2) - 8t + 9 = 0$ $t^2 - 8t + 7 = 0$ $(t-7)(t-1) = 0$ Case $1$: $t = 7$ $\Rightarrow x+\frac{1}{x} = 7$ $\Rightarrow x^2-7x+1 = 0$. The discriminant $D = (-7)^2 - 4(1)(1) = 45 > 0$,so there are two real roots. The product of these roots is $c/a = 1$. Case $2$: $t = 1$ $\Rightarrow x+\frac{1}{x} = 1$ $\Rightarrow x^2-x+1 = 0$. The discriminant $D = (-1)^2 - 4(1)(1) = -3 Thus,the product of all real roots is $1$.

The product of all the real roots of $x^2-8x+9-\frac{8}{x}+\frac{1}{x^2}=0$ is

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