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A preimage of 53 in the function f whose criterion is given by f(x)=x2+4x+8 is

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Answer to a math question A preimage of 53 in the function f whose criterion is given by f(x)=x2+4x+8 is

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Hermann
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125 Answers
1. Empezamos con la función dada:

f(x) = x^2 + 4x + 8

2. Queremos encontrar un valor de x tal que:

f(x) = 53

3. Sustituimos en la función para obtener la ecuación cuadrática:

x^2 + 4x + 8 = 53

4. Restamos 53 en ambos lados para igualar a cero:

x^2 + 4x + 8 - 53 = 0

5. Simplificamos la ecuación:

x^2 + 4x - 45 = 0

6. Resolvemos la ecuación cuadrática usando la fórmula cuadrática:

x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}

Donde a = 1, b = 4 y c = -45.

7. Sustituimos estos valores en la fórmula cuadrática:

x = \frac{{-4 \pm \sqrt{{4^2 - 4(1)(-45)}}}}{2(1)}

8. Simplificamos dentro de la raíz:

x = \frac{{-4 \pm \sqrt{{16 + 180}}}}{2}

9. Continuamos simplificando:

x = \frac{{-4 \pm \sqrt{196}}}{2}

x = \frac{{-4 \pm 14}}{2}

10. Consideramos ambas soluciones para x:

x_1 = \frac{{-4 + 14}}{2} = 5

x_2 = \frac{{-4 - 14}}{2} = -9

11. Las soluciones son x = 5 \text{ o } x = -9

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