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What are the zeros of the quadratic function [tex]$f(x) = 2x^2 + 16x - 9$[/tex]?

A. [tex]$x = -4 - \sqrt{\frac{7}{2}}$[/tex] and [tex][tex]$x = -4 + \sqrt{\frac{7}{2}}$[/tex][/tex]

B. [tex]$x = -4 - \sqrt{\frac{25}{2}}$[/tex] and [tex]$x = -4 + \sqrt{\frac{25}{2}}$[/tex]

C. [tex][tex]$x = -4 - \sqrt{\frac{21}{2}}$[/tex][/tex] and [tex]$x = -4 + \sqrt{\frac{21}{2}}$[/tex]

D. [tex]$x = -4 - \sqrt{\frac{41}{2}}$[/tex] and [tex][tex]$x = -4 + \sqrt{\frac{41}{2}}$[/tex][/tex]


Sagot :

To find the zeros of the quadratic function [tex]\( f(x) = 2x^2 + 16x - 9 \)[/tex], we can use the quadratic formula:

[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]

First, identify the coefficients [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] from the quadratic equation [tex]\( 2x^2 + 16x - 9 = 0 \)[/tex]:

[tex]\[ a = 2, \quad b = 16, \quad c = -9 \][/tex]

Next, calculate the discriminant, which is given by [tex]\( \Delta = b^2 - 4ac \)[/tex]:

[tex]\[ \Delta = 16^2 - 4 \cdot 2 \cdot (-9) = 256 + 72 = 328 \][/tex]

Now, use the quadratic formula to find the roots:

[tex]\[ x_{1,2} = \frac{-16 \pm \sqrt{328}}{2 \cdot 2} = \frac{-16 \pm \sqrt{328}}{4} \][/tex]

Simplify the expression under the square root:

[tex]\[ \sqrt{328} = \sqrt{4 \cdot 82} = 2\sqrt{82} \][/tex]

Substitute [tex]\( \sqrt{328} \)[/tex] into the quadratic formula:

[tex]\[ x_{1,2} = \frac{-16 \pm 2\sqrt{82}}{4} = \frac{-16}{4} \pm \frac{2\sqrt{82}}{4} = -4 \pm \frac{\sqrt{82}}{2} \][/tex]

So, the roots or zeros of the quadratic function are:

[tex]\[ x_1 = -4 - \frac{\sqrt{82}}{2}, \quad x_2 = -4 + \frac{\sqrt{82}}{2} \][/tex]

Now let's see which of the given options match these zeros:

[tex]\[ x = -4 - \sqrt{\frac{41}{2}} \quad \text{and} \quad x = -4 + \sqrt{\frac{41}{2}} \][/tex]

Given that [tex]\(\frac{41}{2} = 20.5\)[/tex] and [tex]\(\sqrt{41/2} = \sqrt{20.5}\)[/tex]:

[tex]\[ -4 - \sqrt{20.5} \quad \text{and} \quad -4 + \sqrt{20.5} \][/tex]

Since [tex]\(\sqrt{82}/2 \approx \sqrt{20.5}\)[/tex], the zeros are:

[tex]\[ x = -4 - \sqrt{\frac{41}{2}} \quad \text{and} \quad x = -4 + \sqrt{\frac{41}{2}} \][/tex]

Thus, the correct matching option is:

[tex]\[ x = -4 - \sqrt{\frac{41}{2}} \quad \text{and} \quad x = -4 + \sqrt{\frac{41}{2}} \][/tex]