Westonci.ca connects you with experts who provide insightful answers to your questions. Join us today and start learning! Get immediate and reliable solutions to your questions from a community of experienced experts on our Q&A platform. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
Sagot :
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].
We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.