Answer :
To determine the zeros of the quadratic function [tex]\( f(x) = 2x^2 + 16x - 9 \)[/tex], we use the quadratic formula. The quadratic formula states:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] are the coefficients of the quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex].
In our case, the coefficients are:
[tex]\[ a = 2, \quad b = 16, \quad c = -9 \][/tex]
First, let's compute the discriminant, [tex]\( \Delta \)[/tex]:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
[tex]\[ \Delta = 16^2 - 4 \cdot 2 \cdot (-9) \][/tex]
[tex]\[ \Delta = 256 + 72 \][/tex]
[tex]\[ \Delta = 328 \][/tex]
Now, we use the quadratic formula to find the roots:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
[tex]\[ x = \frac{-16 \pm \sqrt{328}}{2 \cdot 2} \][/tex]
[tex]\[ x = \frac{-16 \pm \sqrt{328}}{4} \][/tex]
Next, we simplify [tex]\( \sqrt{328} \)[/tex]:
[tex]\[ \sqrt{328} = \sqrt{4 \cdot 82} = 2\sqrt{82} \][/tex]
Substituting this back into the formula, we get:
[tex]\[ x = \frac{-16 \pm 2\sqrt{82}}{4} \][/tex]
[tex]\[ x = \frac{-16}{4} \pm \frac{2\sqrt{82}}{4} \][/tex]
[tex]\[ x = -4 \pm \frac{\sqrt{82}}{2} \][/tex]
To match our solutions with the given options, we convert [tex]\( \sqrt{82} \)[/tex] in another form:
[tex]\[ \frac{\sqrt{82}}{2} = \sqrt{\frac{82}{4}} = \sqrt{\frac{41}{2}} \][/tex]
Thus, the zeros of the quadratic function are:
[tex]\[ x = -4 - \sqrt{\frac{41}{2}} \][/tex]
[tex]\[ x = -4 + \sqrt{\frac{41}{2}} \][/tex]
Hence, the correct option is:
[tex]\[ x = -4 - \sqrt{\frac{41}{2}} \quad \text{and} \quad x = -4 + \sqrt{\frac{41}{2}} \][/tex]
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] are the coefficients of the quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex].
In our case, the coefficients are:
[tex]\[ a = 2, \quad b = 16, \quad c = -9 \][/tex]
First, let's compute the discriminant, [tex]\( \Delta \)[/tex]:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
[tex]\[ \Delta = 16^2 - 4 \cdot 2 \cdot (-9) \][/tex]
[tex]\[ \Delta = 256 + 72 \][/tex]
[tex]\[ \Delta = 328 \][/tex]
Now, we use the quadratic formula to find the roots:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
[tex]\[ x = \frac{-16 \pm \sqrt{328}}{2 \cdot 2} \][/tex]
[tex]\[ x = \frac{-16 \pm \sqrt{328}}{4} \][/tex]
Next, we simplify [tex]\( \sqrt{328} \)[/tex]:
[tex]\[ \sqrt{328} = \sqrt{4 \cdot 82} = 2\sqrt{82} \][/tex]
Substituting this back into the formula, we get:
[tex]\[ x = \frac{-16 \pm 2\sqrt{82}}{4} \][/tex]
[tex]\[ x = \frac{-16}{4} \pm \frac{2\sqrt{82}}{4} \][/tex]
[tex]\[ x = -4 \pm \frac{\sqrt{82}}{2} \][/tex]
To match our solutions with the given options, we convert [tex]\( \sqrt{82} \)[/tex] in another form:
[tex]\[ \frac{\sqrt{82}}{2} = \sqrt{\frac{82}{4}} = \sqrt{\frac{41}{2}} \][/tex]
Thus, the zeros of the quadratic function are:
[tex]\[ x = -4 - \sqrt{\frac{41}{2}} \][/tex]
[tex]\[ x = -4 + \sqrt{\frac{41}{2}} \][/tex]
Hence, the correct option is:
[tex]\[ x = -4 - \sqrt{\frac{41}{2}} \quad \text{and} \quad x = -4 + \sqrt{\frac{41}{2}} \][/tex]
