Data lequazione $$ (k-1) x^{2}-2 k x+k=0 $$, con $$ k eq 1 $$, determina per quali valori di $$ k $$ le soluzioni sono: a. non reali; b. reali ed entrambe positive; c. reali antireciproche; d. reali e tali che il quadrato della somma delle soluzioni sia $$ 9 . \quad $$ a. $$ k1 $$; c. $$ k=\frac{1}{ ext {. d. } k=\frac{3}{6}} $$

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Data lequazione $$ (k-1) x^{2}-2 k x+k=0 $$, con $$ k 
eq 1 $$, determina per quali valori di $$ k $$ le soluzioni sono: a. non reali; b. reali ed entrambe positive; c. reali antireciproche; d. reali e tali che il quadrato della somma delle soluzioni sia $$ 9 . \quad $$ a. $$ k<0 ; $$ b. $$ k>1 $$; c. $$ k=\frac{1}{	ext {. d. } k=\frac{3}{6}} $$

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We start with the quadratic equation $$ (k-1)x^2 - 2kx + k = 0 $$. The coefficients are: - $$ a = k - 1 $$ - $$ b = -2k $$ - $$ c = k $$ We will analyze the conditions for the solutions based on the discriminant $$ D = b^2 - 4ac $$ and the relationships between the roots. For the solutions to be non-real, the discriminant must be less than zero: $$ D = (-2k)^2 - 4(k-1)(k) < 0 $$ This simplifies to: $$ 4k^2 - 4(k^2 - k) < 0 \implies 4k < 0 \implies k < 0 $$ Thus, the condition for non-real solutions is $$ k < 0 $$. For the solutions to be real and both positive, we need: 1. The discriminant must be non-negative: $$ D \geq 0 $$ 2. The sum of the roots $$ S = -\frac{b}{a} = \frac{2k}{k-1} > 0 $$ 3. The product of the roots $$ P = \frac{c}{a} = \frac{k}{k-1} > 0 $$ From $$ D \geq 0 $$: $$ 4k^2 - 4(k^2 - k) \geq 0 \implies 4k \geq 0 \implies k \geq 0 $$ Since $$ k \neq 1 $$, we also need $$ k > 1 $$ for both roots to be positive. Thus, the condition for real and both positive solutions is $$ k > 1 $$. For the solutions to be real and reciprocals, we need: 1. The product of the roots $$ P = 1 $$ $$ \frac{k}{k-1} = 1 \implies k = k - 1 \implies 1 = 0 \text{ (not possible)} $$ Thus, we need to analyze the condition for $$ k $$ such that the roots are reciprocals. The correct condition is $$ k = \frac{1}{1} $$ or $$ k = 1 $$, but since $$ k \neq 1 $$, we cannot have real reciprocal solutions. We need the square of the sum of the solutions to equal 9: $$ S^2 = 9 \implies \left(\frac{2k}{k-1}\right)^2 = 9 \implies \frac{4k^2}{(k-1)^2} = 9 $$ This leads to: $$ 4k^2 = 9(k-1)^2 \implies 4k^2 = 9(k^2 - 2k + 1) \implies 4k^2 = 9k^2 - 18k + 9 $$ Rearranging gives: $$ 5k^2 - 18k + 9 = 0 $$ Now we can solve this quadratic equation for $$ k $$. Solve the quadratic equation by following steps: - step0: Solve by factoring: $$5k^{2}-18k+9=0$$ - step1: Factor the expression: $$\left(k-3\right)\left(5k-3\right)=0$$ - step2: Separate into possible cases: $$\begin{align}&k-3=0\&5k-3=0\end{align}$$ - step3: Solve the equation: $$\begin{align}&k=3\&k=\frac{3}{5}\end{align}$$ - step4: Rewrite: $$k_{1}=\frac{3}{5},k_{2}=3$$ a. Non-reali: $$ k < 0 $$ \ b. Reali ed entrambe positive: $$ k > 1 $$ \ c. Reali antireciproche: Non esistono valori di $$ k $$ validi. \ d. Quadrato della somma delle soluzioni è 9: $$ k = \frac{3}{5} $$ o $$ k = 3 $$.

Data lequazione , con , determina per quali valori di le soluzioni sono:
a. non reali;
b. reali ed entrambe positive;
c. reali antireciproche;
d. reali e tali che il quadrato della somma delle soluzioni sia a. b. ; c.

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Data lequazione , con , determina per quali valori di le soluzioni sono: a. non reali; b. reali ed entrambe positive; c. reali antireciproche; d. reali e tali che il quadrato della somma delle soluzioni sia a. b. ; c.