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masters · #2 November 13th, 2008, 11:43 AM

Quote:

Originally Posted by thestien

hi there im new to this forum and currently teaching myself maths

the problem i have is quadratic equations. the questions im stuck on are complete the square, giving the solution in surd form.

there is an example question i get all of it apart from the last few bits if somebody could point me in the right direction i would appreciate it

here is the example could you explain each step for me please

sqr= square root

2x^2-3x-3= 0
2x^2-3x = 3 (plus 3 to both sides)
2(x^2-3/2) = 3 (eliminate the co eff of x^2)
x^2-3/2x = 3/2 (divide both sides by 2)
x^2- 3/2x +9/16 = 3/2+9/16 (add half of the co eff of x then square it)
(x-3/4)^2 = 33/16 (simplify)
x-3/4 =sqr33/16 or (find the square of both sides)
x-3/4 =-sqr33/16 (ok i understand up to here)

$x-\frac{3}{4}=\pm\sqrt{\frac{33}{16}}$

Quote:

Originally Posted by thestien

x-3/4= 1/4sqr33 (how does sqr33/16 turn into 1/4sqr33 )

$x-\frac{3}{4}=\pm\frac{\sqrt{33}}{\sqrt{16}}$

$x-\frac{3}{4}=\pm\frac{\sqrt{33}}{4}$

$x-\frac{3}{4}=\pm\frac{1}{4}\sqrt{33}$

Quote:

Originally Posted by thestien

(there are 20 questions i have to solve in my book in this format but i cant work out how to manipulate the sqr like that?)
x=3/4+1/4sqr33
x=3/4-1/4sqr33

thank you in advance im sure its a simple rule that i havent learned yet

$x=\frac{3}{4}\pm\frac{1}{4}\sqrt{33}$

$x=\frac{3\pm\sqrt{33}}{4}$