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Two times the square of a number, added to twelve times the number, equals -1. Find all such numbers.

A. [tex] -3+\sqrt{34}, -3-\sqrt{34} [/tex]
B. [tex] \frac{-6+\sqrt{34}}{4}, \frac{-6-\sqrt{34}}{4} [/tex]
C. [tex] \frac{-12+\sqrt{34}}{2}, \frac{-12-\sqrt{34}}{2} [/tex]
D. [tex] \frac{-6+\sqrt{34}}{2}, \frac{-6-\sqrt{34}}{2} [/tex]



Answer :

GinnyAnswer
Let's rewrite and solve the given problem step-by-step.

The problem describes a quadratic equation:

[tex]\[ 2x^2 + 12x + 1 = 0 \][/tex]

Our goal is to find the values of [tex]\( x \)[/tex] that satisfy this equation.

The standard form of a quadratic equation is:

[tex]\[ ax^2 + bx + c = 0 \][/tex]

For our specific equation:

[tex]\[ a = 2 \][/tex]
[tex]\[ b = 12 \][/tex]
[tex]\[ c = 1 \][/tex]

We can use the quadratic formula to find the solutions for [tex]\( x \)[/tex]:

[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]

1. Calculate the discriminant ([tex]\( \Delta \)[/tex]):

[tex]\[ \Delta = b^2 - 4ac \][/tex]
[tex]\[ \Delta = 12^2 - 4 \cdot 2 \cdot 1 \][/tex]
[tex]\[ \Delta = 144 - 8 \][/tex]
[tex]\[ \Delta = 136 \][/tex]

2. Calculate the roots using the quadratic formula:

[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
[tex]\[ x = \frac{-12 \pm \sqrt{136}}{2 \cdot 2} \][/tex]
[tex]\[ x = \frac{-12 \pm \sqrt{136}}{4} \][/tex]

Simplifying the roots, we find:

[tex]\[ x = \frac{-6 \pm \sqrt{34}}{2} \][/tex]

So, the solutions for the equation [tex]\( 2x^2 + 12x + 1 = 0 \)[/tex] are:

[tex]\[ x = \frac{-6 + \sqrt{34}}{2} \quad \text{and} \quad x = \frac{-6 - \sqrt{34}}{2} \][/tex]

From the given choices, the correct answer is:

[tex]\[ \boxed{\frac{-6+\sqrt{34}}{2}, \frac{-6-\sqrt{34}}{2}} \][/tex]

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