Solving Equations be the Quadratic Formula

[mogfog]

Hey guys, i need some real help with this one, i can't figure it out for the life of me. it is for a summer packet, just review, and i don't get it. Thequestions are...

(leave in simplified radical form when possible.)


1. 3x2 + 4x - 3 = 0


PLease help!

Thanks,
~grant

 

[Loren]

3x^2 + 4x - 3 = 0

a=3, b=4, c=-3

Plug these values into...
$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
and do the simplification.

 

[vannahbabyx3]

Quadratic Formula:
-b (+/-) ? (b[sup:2vp1ixbt]2[/sup:2vp1ixbt]-4ac)/2a

Plug in your original equation. -->

a = 3
b = 4
c = -3

-4 (+/-) ? (4[sup:2vp1ixbt]2[/sup:2vp1ixbt]-4(3)(-3))/2(3)

Solve what's under the radical, and simplify where possible. --> -4 (+/-) ? (52)/6

Since it was not a perfect square, I left the product under the radical.

My answers are:
-2/3 + ? (52)/6 and -2/3 - ? (52)/6

Hope that was helpful. =)

 

[stapel]

Quadratic Formula:
-b (+/-) ? (b[sup:3b1nky1y]2[/sup:3b1nky1y]-4ac)/2a

Plug in your original equation. -->

Do you perhaps mean "plug the coefficients from the equation into the Quadratic Formula"...? (The "equation" does not, of course, get "plugged in" anywhere.)

A tutor already explained what the Formula was (providing an accurate statement of it), but it should perhaps be mentioned that the Formula may only be applied once the quadratic equation is "solved" for a decreasing polynomial on one side of the "equals" sign, and "zero" on the other. Before this point, the Formula does not apply. :!:

a = 3
b = 4
c = -3

-4 (+/-) ? (4[sup:3b1nky1y]2[/sup:3b1nky1y]-4(3)(-3))/2(3)

Do you perhaps mean the following...?

. . . . .$\displaystyle \frac{-4\, \pm\, \sqrt{4^2\, -\, 4(3)(-3)}}{2(3)}$

(Your formatting leads to incorrect values; the above formatting is necessary to arrive at the solution.)

To the original poster: Please reply if you are confused, or if you need clarification of some particular point. For instance, are you allowed to use the Quadratic Formula for this exercise, or are you supposed to find the solution in some other manner? (Your subject line is a tad unclear.) :?:

Thank you!

Eliz.