Answer:
The zeros to the quadratic equation are:
[tex]x= -4+\sqrt{\frac{41}{2}}\\\\x= -4-\sqrt{\frac{41}{2}}[/tex]
Step-by-step explanation:
A quadratic function is one of the form [tex]f(x) = ax^2 + bx + c[/tex], 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 [tex]f(x) = 0[/tex] or [tex]ax^2 + bx +c = 0[/tex].
To find the zeros of the quadratic function [tex]f(x)= 2x^2 + 16x -9[/tex] , we set [tex]f(x) = 0[/tex], and solve the equation.
[tex]2x^2+16x\:-9=0[/tex]
[tex]\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}[/tex]
[tex]\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}}[/tex]
