Question

What are the zeros of the quadratic function f(x) = 2x2 + 8x – 3?

Answer 1

Answer:

$\large\boxed{x=-2-\sqrt{\dfrac{11}{2}}\ \text{and}\ x=-2+\sqrt{\dfrac{11}{2}}}$

Step-by-step explanation:

$f(x)=2x^2+8x-3\\\\\text{You can use the quadratic formula:}\\\\ax^2+bx+c=0\\\\x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\\text{We have}\ a=2,\ b=8,\ c=-3\\\\\text{Substitute:}\\\\b^2-4ac=8^2-4(2)(-3)=64+24=88\\\\\sqrt{b^2-4ac}=\sqrt{88}=\sqrt{4\cdot22}=\sqrt4\cdot\sqrt{22}=2\sqrt{22}\\\\x=\dfrac{-8\pm2\sqrt{22}}{2(2)}=\dfrac{-8}{4}\pm\dfrac{2\sqrt{22}}{4}=-2\pm\dfrac{\sqrt{22}}{2}=-2\pm\dfrac{\sqrt{22}}{\sqrt4}\\\\x=-2\pm\sqrt{\dfrac{22}{4}}=-2\pm\sqrt{\dfrac{11}{2}}$

Answer 2

Answer:

ANSWER IS A!!!!!!!!!!!!

Step-by-step explanation: