## anonymous 5 years ago Write quadratic equation in standard form with given solution set: {1/6, -2/3}

1. anonymous

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

2. anonymous

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.

3. anonymous

And then what? Use the quadratic formula?

4. anonymous

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"

5. anonymous

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

6. anonymous

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

7. anonymous

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

8. anonymous

Awesome, thank you so much.

9. anonymous

welcome, but check my algebra, i am tired.

10. anonymous

It is correct.