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]
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]
