At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
Do you mean quadratic? Use the formula and plug in the coeff of the x^2 for a, the coeff of the x term for b and the coeff of the constant term for c. 2x^2+13x=4 ---> 2x^2+13x-4 2x^2--->a=2 13x------>b=13 -4---->c=-4
-b + or - sqrt(b^2-4ac) all divided by 2a
Hold on i have chocies
it would be better to say plus and minus, you get two answers
they should both work as solutions to the equation
If you're doing homework they probably want both answers.
A.) x= 13+√201 -------------- 2 B.)x=-13+√177 ---------------- 4 C.) x= -13+√ 201 ---------------- 2 D.) x= -13 +√ 201 ----------- 4
AND + HAS a line under it \[\pm\]
Close, not quite.
substitute your values in the quadratic formula and see which is the correct one.
what's the difference between C and D
c has a 2 and d has a 4
Well the answer is exactly the same as D, only it's + and - sqrt201
katie, there is no point guessing the answer. take some time and substitute your vlues into the quadratic equation. that will give you the correct answer.
so its D?
If the + is a + - line thing, yes
yes it has this \[\pm\] under it so its D
Yeah, i mean it's exactly the same as the formula, it just matters that you plug in the right numbers.
Check your work on http://www.wolframalpha.com/input/?i=2x^2%2B13x%3D4+quadratic+formula
I've gotta go, goodluck