Question

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

Answer 1

$2x^{2} +16x - 9 = 0 2(x + 4)^{2} - 41 = 0 (x + 4)^{2} = \frac{41}{2} x + 4 = \pm \sqrt{\frac{41}{2}} x = -4 \pm \sqrt{\frac{41}{2}}$

Answer 2

Answer:

The zeros to the quadratic equation are:

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

Step-by-step explanation:

A quadratic function is one of the form $f(x) = ax^2 + bx + c$, where a, b, and c are numbers with a not equal to zero.

The zeros of a quadratic function are the two values of x when $f(x) = 0$ or $ax^2 + bx +c = 0$.

To find the zeros of the quadratic function $f(x)= 2x^2 + 16x -9$ , we set $f(x) = 0$, and solve the equation.

$2x^2+16x\:-9=0$

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=2,\:b=16,\:c=-9:\quad x_{1,\:2}=\frac{-16\pm \sqrt{16^2-4\cdot \:2\left(-9\right)}}{2\cdot \:2}\\\\x=\frac{-16+\sqrt{16^2-4\cdot \:2\left(-9\right)}}{2\cdot \:2}= -4+\sqrt{\frac{41}{2}}\\\\x=\frac{-16-\sqrt{16^2-4\cdot \:2\left(-9\right)}}{2\cdot \:2}= -4-\sqrt{\frac{41}{2}}$