Answer :
To find the zeros of the quadratic function [tex]\( f(x) = 2x^2 + 16x - 9 \)[/tex], we need to solve for [tex]\( x \)[/tex] in the equation [tex]\( 2x^2 + 16x - 9 = 0 \)[/tex].
To solve this quadratic equation, we can use the quadratic formula which is given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For the quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex], the coefficients are:
- [tex]\( a = 2 \)[/tex]
- [tex]\( b = 16 \)[/tex]
- [tex]\( c = -9 \)[/tex]
First, we calculate the discriminant [tex]\(\Delta\)[/tex]:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Plugging in the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[ \Delta = (16)^2 - 4 \cdot 2 \cdot (-9) = 256 + 72 = 328 \][/tex]
Now, we substitute [tex]\(\Delta\)[/tex] back into the quadratic formula:
[tex]\[ x = \frac{-16 \pm \sqrt{328}}{4} \][/tex]
Next, we simplify [tex]\(\sqrt{328}\)[/tex]:
[tex]\[ \sqrt{328} = \sqrt{4 \times 82} = 2\sqrt{82} \][/tex]
So the expression for [tex]\( x \)[/tex] becomes:
[tex]\[ x = \frac{-16 \pm 2\sqrt{82}}{4} \][/tex]
Dividing both terms in the numerator by 4:
[tex]\[ x = \frac{-16}{4} \pm \frac{2\sqrt{82}}{4} \][/tex]
[tex]\[ x = -4 \pm \frac{\sqrt{82}}{2} \][/tex]
Next, to align our answer with the given options, observe that [tex]\(\sqrt{82}\)[/tex] can be written as:
[tex]\[ \sqrt{82} = \sqrt{41 \cdot 2} = \sqrt{41 \cdot 2} = \sqrt{\frac{41}{2} \cdot 4} = 2\sqrt{\frac{41}{2}} \][/tex]
Thus, the expression for [tex]\( x \)[/tex] becomes:
[tex]\[ x = -4 \pm \sqrt{\frac{41}{2}} \][/tex]
Therefore, the zeros of the quadratic function [tex]\( 2x^2 + 16x - 9 \)[/tex] are:
[tex]\[ x = -4 - \sqrt{\frac{41}{2}} \quad \text{and} \quad x = -4 + \sqrt{\frac{41}{2}} \][/tex]
Thus, the correct option is:
[tex]\( x = -4 - \sqrt{\frac{41}{2}} \)[/tex] and [tex]\( x = -4 + \sqrt{\frac{41}{2}} \)[/tex].
To solve this quadratic equation, we can use the quadratic formula which is given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For the quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex], the coefficients are:
- [tex]\( a = 2 \)[/tex]
- [tex]\( b = 16 \)[/tex]
- [tex]\( c = -9 \)[/tex]
First, we calculate the discriminant [tex]\(\Delta\)[/tex]:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Plugging in the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[ \Delta = (16)^2 - 4 \cdot 2 \cdot (-9) = 256 + 72 = 328 \][/tex]
Now, we substitute [tex]\(\Delta\)[/tex] back into the quadratic formula:
[tex]\[ x = \frac{-16 \pm \sqrt{328}}{4} \][/tex]
Next, we simplify [tex]\(\sqrt{328}\)[/tex]:
[tex]\[ \sqrt{328} = \sqrt{4 \times 82} = 2\sqrt{82} \][/tex]
So the expression for [tex]\( x \)[/tex] becomes:
[tex]\[ x = \frac{-16 \pm 2\sqrt{82}}{4} \][/tex]
Dividing both terms in the numerator by 4:
[tex]\[ x = \frac{-16}{4} \pm \frac{2\sqrt{82}}{4} \][/tex]
[tex]\[ x = -4 \pm \frac{\sqrt{82}}{2} \][/tex]
Next, to align our answer with the given options, observe that [tex]\(\sqrt{82}\)[/tex] can be written as:
[tex]\[ \sqrt{82} = \sqrt{41 \cdot 2} = \sqrt{41 \cdot 2} = \sqrt{\frac{41}{2} \cdot 4} = 2\sqrt{\frac{41}{2}} \][/tex]
Thus, the expression for [tex]\( x \)[/tex] becomes:
[tex]\[ x = -4 \pm \sqrt{\frac{41}{2}} \][/tex]
Therefore, the zeros of the quadratic function [tex]\( 2x^2 + 16x - 9 \)[/tex] are:
[tex]\[ x = -4 - \sqrt{\frac{41}{2}} \quad \text{and} \quad x = -4 + \sqrt{\frac{41}{2}} \][/tex]
Thus, the correct option is:
[tex]\( x = -4 - \sqrt{\frac{41}{2}} \)[/tex] and [tex]\( x = -4 + \sqrt{\frac{41}{2}} \)[/tex].
