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][tex]$x=-1+\sqrt{\frac{13}{6}}$[/tex][/tex]

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

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

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



Answer :

To determine 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]. The general form of a quadratic equation is [tex]\( ax^2 + bx + c = 0 \)[/tex], where [tex]\( a = 6 \)[/tex], [tex]\( b = 12 \)[/tex], and [tex]\( c = -7 \)[/tex] in this case.

The quadratic formula is given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]

Step-by-step:

1. Identify the coefficients:
[tex]\[ a = 6, \quad b = 12, \quad c = -7 \][/tex]

2. Calculate the discriminant ([tex]\( \Delta \)[/tex]):
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Plugging in the values:
[tex]\[ \Delta = 12^2 - 4 \cdot 6 \cdot (-7) \][/tex]
[tex]\[ \Delta = 144 + 168 \][/tex]
[tex]\[ \Delta = 312 \][/tex]

3. Apply the quadratic formula:
[tex]\[ x = \frac{-12 \pm \sqrt{312}}{12} \][/tex]

4. Simplify the expression inside the square root:
[tex]\[ \sqrt{312} = 2 \times \sqrt{78} = 2 \times \sqrt{6 \times 13} = 2 \sqrt{78} \][/tex]

5. Substitute [tex]\(\sqrt{312}\)[/tex] back into the quadratic formula:
[tex]\[ x = \frac{-12 \pm 2\sqrt{78}}{12} \][/tex]

6. Simplify the fraction:
[tex]\[ x = \frac{-12}{12} \pm \frac{2\sqrt{78}}{12} \][/tex]
[tex]\[ x = -1 \pm \frac{\sqrt{78}}{6} \][/tex]

Since [tex]\(\sqrt{78}\)[/tex] simplifies to [tex]\( \sqrt{6 \times 13}\)[/tex]:

[tex]\[ x = -1 \pm \sqrt{\frac{78}{36}} = -1 \pm \sqrt{\frac{13}{6}} \][/tex]

So, the zeros of the quadratic function are:
[tex]\[ x = -1 - \sqrt{\frac{13}{6}} \quad \text{and} \quad x = -1 + \sqrt{\frac{13}{6}} \][/tex]

This matches the first option provided. Thus, the correct answer is:
[tex]\[ \boxed{x = -1 - \sqrt{\frac{13}{6}} \quad \text{and} \quad x = -1 + \sqrt{\frac{13}{6}}} \][/tex]