Join IDNLearn.com to access a wealth of knowledge and get your questions answered by experts. Get step-by-step guidance for all your technical questions from our knowledgeable community members.

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].