1. ## Quadratic Equations, Exam Tommorow.

I have a few problems that I'm trying to understand. If anyone can assist me that would be greatly appreciated.

Solve by using square root property:
1) 3(x-4)^2 = 15
2) (3x-4)^2 = 8

Solve by completing the square:
3) 2x^2 - 7x + 3= 0

4) 3x^2 - 3x - 4= 0

Thank you.

2. Originally Posted by FactoringAnguish
I have a few problems that I'm trying to understand. If anyone can assist me that would be greatly appreciated.

Solve by using square root property:
1) 3(x-4)^2 = 15
2) (3x-4)^2 = 8
i'll do 1, do number 2..

$3(x-4)^2 = 15 \implies (x-4)^2 = 5 \implies \sqrt {(x-4)^2} = \sqrt 5$

$\implies |x-4| = \sqrt 5 \implies x - 4 = \pm \sqrt 5 \implies x = 4 \pm \sqrt 5$

Originally Posted by FactoringAnguish

Solve by completing the square:
3) 2x^2 - 7x + 3= 0
$2x^2 - 7x + 3 = 0 \implies x^2 - \frac{7}{2}x + \frac{3}{2} = 0$

$\implies x^2 - \frac{7}{2}x = - \frac{3}{2}$ adding $\frac{49}{16}$ sides, we have $x^2 - \frac{7}{2}x + \frac{49}{16} = \frac{49}{16} - \frac{3}{2}$

$\implies \left( {x-\frac{7}{4}} \right) ^2 = \frac{25}{16} \implies \left( {x-\frac{7}{4}} \right) = \pm \frac{5}{4}$

from here, continue.. just like in number 1.

Originally Posted by FactoringAnguish
4) 3x^2 - 3x - 4= 0

Thank you.
if we have a quadratic poly of the form $ax^2 + bx + c = 0$, then $x= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

so, a= 3, b= -3, c = -4.. do the substitution.. Ü

3. Thanks, makes perfect sense now that I see it worked out

4. ## Methods for solving quadratics

Originally Posted by FactoringAnguish
I have a few problems that I'm trying to understand. If anyone can assist me that would be greatly appreciated.

Solve by using square root property:
1) 3(x-4)^2 = 15
2) (3x-4)^2 = 8

Solve by completing the square:
3) 2x^2 - 7x + 3= 0