• Jan 5th 2013, 06:43 AM
Cameron
I have tried the quadratic formula and also completing the square(sorta).

(questions are not in standard form yet)

6x2 = 4 - 5x

4x2 + 2x = 5

9(x - 4)2 + 1 = 0
• Jan 5th 2013, 07:23 AM
Soroban
Hello, Cameron!

Quote:

I have tried the Quadratic Formula and also completing the square (sorta).
If you showed some work, we might find your errors.

(questions are not in standard form yet)
Did you put them in standard form?

$\displaystyle [1]\;6x^2 \:=\: 4 - 5x$

We have: .$\displaystyle 6x^2 + 5x - 4 \:=\:0$

Factor: .$\displaystyle (2x-1)(3x+4) \:=\:0$

Therefore: .$\displaystyle x \:=\:\tfrac{1}{2},\;\text{-}\tfrac{4}{3}$

Quote:

$\displaystyle [2]\;4x^2+ 2x \:=\:5$

We have: .$\displaystyle 4x^2 + 2x - 5 \:=\:0$

Quadratic Formula: .$\displaystyle x \;=\;\frac{\text{-}2 \pm \sqrt{2^2 - 4(4)(\text{-}5)}}{2(4)}$

. . . . . . $\displaystyle x \;=\;\frac{\text{-}2\pm\sqrt{84}}{8} \;=\;\frac{\text{-}2\pm2\sqrt{21}}{8} \;=\;\frac{\text{-}1\pm\sqrt{21}}{4}$

Quote:

$\displaystyle [3]\;9(x - 4)^2 + 1 \:=\: 0$

No fancy tricks . . . just solve for $\displaystyle x.$

We have: .$\displaystyle 9(x-4)^2 \:=\:\text{-}1$

. . . . . . . . . $\displaystyle (x-4)^2 \:=\:\text{-}\tfrac{1}{9}{$

n . . . . . . . . . $\displaystyle x-4 \;=\;\pm\sqrt{\text{-}\tfrac{1}{9}}$

n . . . . . . . . . $\displaystyle x-4 \;=\;\pm\tfrac{1}{3}i$

n . . . . . . . . . . . .$\displaystyle x \;=\;4\pm\tfrac{1}{3}i$