IDNLearn.com: Where your questions are met with thoughtful and precise answers. Join our knowledgeable community and get detailed, reliable answers to all your questions.

What are the zeros of the quadratic function [tex]\( f(x) = 6x^2 + 12x - 7 \)[/tex]?

A. [tex]\( x = -1 - \sqrt{\frac{13}{6}} \)[/tex] and [tex]\( x = -1 + \sqrt{\frac{13}{6}} \)[/tex]

B. [tex]\( x = -1 - \frac{2}{\sqrt{3}} \)[/tex] and [tex]\( x = -1 + \frac{2}{\sqrt{3}} \)[/tex]

C. [tex]\( x = -1 - \sqrt{\frac{7}{6}} \)[/tex] and [tex]\( x = -1 + \sqrt{\frac{7}{6}} \)[/tex]

D. [tex]\( x = -1 - \frac{1}{\sqrt{6}} \)[/tex] and [tex]\( x = -1 + \frac{1}{\sqrt{6}} \)[/tex]


Sagot :

To find the zeros of the quadratic function [tex]\( f(x) = 6x^2 + 12x - 7 \)[/tex], we need to solve the equation [tex]\( 6x^2 + 12x - 7 = 0 \)[/tex].

Here is the step-by-step solution:

1. Identify the coefficients:
We can identify the coefficients from the quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex].
- [tex]\( a = 6 \)[/tex]
- [tex]\( b = 12 \)[/tex]
- [tex]\( c = -7 \)[/tex]

2. Calculate the discriminant:
The discriminant, [tex]\( \Delta \)[/tex], is given by the formula [tex]\( \Delta = b^2 - 4ac \)[/tex].

[tex]\[ \Delta = 12^2 - 4 \cdot 6 \cdot (-7) = 144 + 168 = 312 \][/tex]

3. Find the square root of the discriminant:
To find the roots, we need the square root of the discriminant [tex]\( \sqrt{312} \)[/tex]. For simplicity, we can write:

[tex]\[ \sqrt{312} = \sqrt{4 \times 78} = 2\sqrt{78} \][/tex]

4. Apply the quadratic formula:
The quadratic formula is:

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

Substituting in the values we have:

[tex]\[ x = \frac{-12 \pm 2\sqrt{78}}{2 \cdot 6} = \frac{-12 \pm 2\sqrt{78}}{12} \][/tex]

5. Simplify the expression:
Dividing the numerator by the denominator:

[tex]\[ x = \frac{-12}{12} \pm \frac{2\sqrt{78}}{12} \][/tex]

Simplifying both parts:

[tex]\[ x = -1 \pm \frac{\sqrt{78}}{6} \][/tex]

6. Write the solutions:
Therefore, the two solutions are:

[tex]\[ x_1 = -1 - \frac{\sqrt{78}}{6} \][/tex]
[tex]\[ x_2 = -1 + \frac{\sqrt{78}}{6} \][/tex]

So the zeros of the quadratic function [tex]\( f(x) = 6x^2 + 12x - 7 \)[/tex] are:

- [tex]\( x = -1 - \frac{\sqrt{78}}{6} \)[/tex]
- [tex]\( x = -1 + \frac{\sqrt{78}}{6} \)[/tex]

These solutions correspond to:

[tex]\[ x = -1 - \frac{\sqrt{78}}{6} \][/tex]
[tex]\[ x = -1 + \frac{\sqrt{78}}{6} \][/tex]

Note that among the given choices, none of them directly match this simplified form. However, we can conclude that the given correct roots are: [tex]\( x = -1 - \frac{\sqrt{78}}{6} \)[/tex] and [tex]\( x = -1 + \frac{\sqrt{78}}{6} \)[/tex].