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Find the value(s) of k if the quadratic equation3 3 4 02x k x− + = has real roots.

Question

Find the value(s) of k if the quadratic equation3 3 4 02x k x− + = has real roots.

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Solution

Para que la ecuación cuadrática 2x23kx+4=02x^2 - 3kx + 4 = 0 tenga raíces reales, el discriminante debe ser mayor o igual a cero. El discriminante de una ecuación cuadrática ax2+bx+c=0ax^2 + bx + c = 0 está dado por b24acb^2 - 4ac.

En este caso, a=2a = 2, b=3kb = -3k, y c=4c = 4. Entonces, el discriminante es:

(3k)2424 (-3k)^2 - 4 \cdot 2 \cdot 4

Simplificamos:

9k232 9k^2 - 32

Para que la ecuación tenga raíces reales, el discriminante debe ser mayor o igual a cero:

9k2320 9k^2 - 32 \geq 0

Resolvemos esta desigualdad:

9k232 9k^2 \geq 32

Dividimos ambos lados por 9:

k2329 k^2 \geq \frac{32}{9}

Tomamos la raíz cuadrada de ambos lados:

k329 |k| \geq \sqrt{\frac{32}{9}}

Simplificamos la raíz cuadrada:

k323 |k| \geq \frac{\sqrt{32}}{3}

Sabemos que 32=162=42\sqrt{32} = \sqrt{16 \cdot 2} = 4\sqrt{2}, entonces:

k423 |k| \geq \frac{4\sqrt{2}}{3}

Esto significa que kk debe ser mayor o igual a 423\frac{4\sqrt{2}}{3} o menor o igual a 423-\frac{4\sqrt{2}}{3}. Por lo tanto, los valores de kk que satisfacen la condición son:

k423ok423 k \leq -\frac{4\sqrt{2}}{3} \quad \text{o} \quad k \geq \frac{4\sqrt{2}}{3}

This problem has been solved

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