Question

What are the zeros of the quadratic function f(x) = 2x2 + 16x – 9?

Answer 1

we have

$f(x)=2x^{2}+16x-9$

To find the zeros equate the function to zero

$2x^{2}+16x-9=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$2x^{2}+16x=9$

Factor the leading coefficient

$2(x^{2}+8x)=9$

Complete the square. Remember to balance the equation by adding the same constants to each side

$2(x^{2}+8x+16)=9+32$

$2(x^{2}+8x+16)=41$

Rewrite as perfect squares

$2(x+4)^{2}=41$

$(x+4)^{2}=(41/2)$

Square root both sides

$x+4=(+/-)\sqrt{\frac{41}{2}}$

$x=-4(+/-)\sqrt{\frac{41}{2}}$

therefore

the answer is

The zeros of the function are

$x=-4+\sqrt{\frac{41}{2}}$

$x=-4-\sqrt{\frac{41}{2}}$

Answer 2

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

Zeros of a function

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.

In summary, the roots or zeros of the quadratic function are those values ​​of x for which the expression is equal to 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:

$x1,x2=\frac{-b+-\sqrt{b^{2}-4ac } }{2a}$

This case

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

Being:

• a= 2
• b=16
• c=-9

the zeros or roots are calculated as:

$x1=\frac{-16+\sqrt{16^{2}-4x2x(-9) } }{2x2}$

$x1=\frac{-16+\sqrt{256+72 } }{4}$

$x1=\frac{-16+\sqrt{328 } }{4}$

$x1=\frac{-16+\sqrt{4x82 } }{4}$

$x1=\frac{-16+2\sqrt{82 } }{4}$

$x1=\frac{-16}{4} +\frac{2\sqrt{82 } }{4}$

$x1=-4+\frac{\sqrt{82 } }{2}$

and

$x2=\frac{-16-\sqrt{16^{2}-4x2x(-9) } }{2x2}$

$x2=\frac{-16-\sqrt{256+72 } }{4}$

$x2=\frac{-16-\sqrt{328 } }{4}$

$x2=\frac{-16-\sqrt{4x82 } }{4}$

$x2=\frac{-16-2\sqrt{82 } }{4}$

$x2=\frac{-16}{4} -\frac{2\sqrt{82 } }{4}$

$x2=-4-\frac{\sqrt{82 } }{2}$

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

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