To find the zeros of the quadratic function [tex]\( f(x) = 6x^2 + 12x - 7 \)[/tex], we can use the quadratic formula, which is given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, the coefficients are [tex]\( a = 6 \)[/tex], [tex]\( b = 12 \)[/tex], and [tex]\( c = -7 \)[/tex]. Let's go through the steps to solve for [tex]\( x \)[/tex].
1. Calculate the Discriminant:
The discriminant ([tex]\( \Delta \)[/tex]) is given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Plugging in the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[ \Delta = 12^2 - 4 \cdot 6 \cdot (-7) \][/tex]
[tex]\[ \Delta = 144 + 168 \][/tex]
[tex]\[ \Delta = 312 \][/tex]
2. Simplify the Discriminant:
The square root of the discriminant is:
[tex]\[ \sqrt{\Delta} = \sqrt{312} \][/tex]
Notice that:
[tex]\[ \sqrt{312} = \sqrt{4 \times 78} = 2\sqrt{78} \][/tex]
3. Apply the Quadratic Formula:
We then substitute the values into the quadratic formula:
[tex]\[ x = \frac{-12 \pm 2\sqrt{78}}{2 \cdot 6} \][/tex]
Simplify the expression:
[tex]\[ x = \frac{-12 \pm 2\sqrt{78}}{12} \][/tex]
[tex]\[ x = \frac{-12}{12} \pm \frac{2\sqrt{78}}{12} \][/tex]
[tex]\[ x = -1 \pm \frac{\sqrt{78}}{6} \][/tex]
Notice that:
[tex]\[ \sqrt{78} = \sqrt{3 \times 26} = \sqrt{3} \times \sqrt{26} \][/tex]
By simplifying [tex]\(\sqrt{78}\)[/tex] more directly, we still have:
[tex]\[ \frac{\sqrt{78}}{6} \][/tex]
4. Compare with Given Options:
Let's rationalize the original exact form given:
[tex]\[ \frac{\sqrt{78}}{6} = \frac{\sqrt{13 \times 6}}{6} = \frac{\sqrt{13}}{\sqrt{6}} = \sqrt{\frac{13}{6}} \][/tex]
Thus, the solutions are:
[tex]\[ x = -1 - \sqrt{\frac{13}{6}} \][/tex]
[tex]\[ x = -1 + \sqrt{\frac{13}{6}} \][/tex]
Therefore, the correct answer is:
[tex]\[ x = -1 - \sqrt{\frac{13}{6}} \quad \text{and} \quad x = -1 + \sqrt{\frac{13}{6}} \][/tex]
