What are the zeros of the quadratic function f(x) = 2x2 + 16x – 9?
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and x = -4+
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2
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O x=-4-
25
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and x = -4+
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V2
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O x=-4-
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and x = -4 +
21
V2
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O x=-4-
41
and x = -4 +
V 2

Answer:

[tex]\mathrm{X\:Intercepts}:\:\left(\frac{-8+\sqrt{82}}{2},\:0\right),\:\left(-\frac{8+\sqrt{82}}{2},\:0\right)[/tex]

Step-by-step explanation:

[tex]f\left(x\right)\:=\:2x^2\:+\:16x\:-\:9[/tex]

- Given

[tex]\mathrm{X\:Intercepts}:\:\left(\frac{-8+\sqrt{82}}{2},\:0\right),\:\left(-\frac{8+\sqrt{82}}{2},\:0\right)[/tex]

By definition of zeros of  a function,  the zeros of the quadratic function f(x) = 2x² + 16x – 9 are [tex]x1=-4+\frac{\sqrt{82}}{2}[/tex] and [tex]x2=-4-\frac{\sqrt{82}}{2}[/tex] .

The points where a polynomial function crosses the axis of the independent term (x) represent the so-called zeros of the function.

That is, the zeros represent the roots of the polynomial equation that is obtained by making f(x)=0.

Graphically, the roots correspond to the abscissa of the points where the parabola intersects the x-axis.

In a quadratic function that has the form:

f(x)= ax² + bx + c

the zeros or roots are calculated by:

[tex]x1,x2=\frac{-b+-\sqrt{b^{2}-4ab } }{2a}[/tex]

The quadratic function is f(x) = 2x² + 16x – 9

Being:

the zeros or roots are calculated as:

[tex]x1=\frac{-16+\sqrt{16^{2}-4x2x(-9) } }{2x2}[/tex]

[tex]x1=\frac{-16+\sqrt{256 +72 } }{4}[/tex]

[tex]x1=\frac{-16+\sqrt{328} }{4}[/tex]

[tex]x1=\frac{-16+\sqrt{4x82} }{4}[/tex]

[tex]x1=\frac{-16+2\sqrt{82} }{4}[/tex]

[tex]x1=\frac{-16}{4}+\frac{2\sqrt{82}}{4}[/tex]

[tex]x1=-4+\frac{\sqrt{82}}{2}[/tex]

and

[tex]x2=\frac{-16-\sqrt{16^{2}-4x2x(-9) } }{2x2}[/tex]

[tex]x2=\frac{-16-\sqrt{256 +72 } }{4}[/tex]

[tex]x2=\frac{-16-\sqrt{328} }{4}[/tex]

[tex]x2=\frac{-16-\sqrt{4x82} }{4}[/tex]

[tex]x2=\frac{-16-2\sqrt{82} }{4}[/tex]

[tex]x2=\frac{-16}{4}-\frac{2\sqrt{82}}{4}[/tex]

[tex]x2=-4-\frac{\sqrt{82}}{2}[/tex]

Finally, the zeros of the quadratic function f(x) = 2x² + 16x – 9 are [tex]x1=-4+\frac{\sqrt{82}}{2}[/tex] and [tex]x2=-4-\frac{\sqrt{82}}{2}[/tex] .

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