Write quadratic equation in standard form with given solution set:
{1/6, -2/3}

Question

anonymous · Answer

since 1/6 is a zero on factor is $$x-\frac{1}{6}$$
another is $$x+\frac{2}{3}$$ so you could write 
$$(x-\frac{1}{6})(x+\frac{2}{3})$$

anonymous · Answer

but instead you would probably write
$$(6x-1)(3x+2)$$
standard form means multiply this out to get 
$$18x^2+9x-2$$ if my algebra is correct.

anonymous · Answer

And then what? Use the quadratic formula?

anonymous · Answer

no.  you already have the zeros.  you were given that the zeros are $$\frac{1}{6}$$ and $$-\frac{2}{3}$$
so you are not being asked to find the zeros.  this question is asked in reverse.  not "here is the equation, find the zeros" but rather "here are the zeros, find the equation"

anonymous · Answer

So the equation would be my final answer then, right?

anonymous · Answer

if you are given that the zeros are a and b then the equation must have been $$(x-a)(x-b)=0$$

anonymous · Answer

yes the equation is what you were asked for.  if i multiplied out correctly, your answer should be $$18x^2+9x-2=0$$

anonymous · Answer

Awesome, thank you so much.

anonymous · Answer

welcome, but check my algebra, i am tired.

anonymous · Answer

It is correct.