Quadratics Help!

**Cameron** · January 5th 2013, 06:43 AM

I have tried the quadratic formula and also completing the square(sorta).
Im Stuck Please help.

(questions are not in standard form yet)

6x²= 4 - 5x

4x²+ 2x = 5

9(x - 4)² + 1 = 0

**Soroban** · January 5th 2013, 07:23 AM

Hello, Cameron!

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

I'm stuck. Please help.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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