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Anyone any good at compelting the sqauer i need help!!!

Plz help me i keep practising completing the sqaure on euqation with coffecients (ax^2+bx+c) infront of the X^2 and negative coeffcients but i keep having to change the way i do it, does anyone know a genereal easy method or rule for working out completing the sqare with coeffeicient plz help im am extremley stressed!!! For example can you take me through THIS: Express
3x^2-8x-3=0 in completed sqaure form, and use your result to solve the equation ?

(Do they get any harder then this, i can do the normal ones no probs but these are confusing me!)

Reply 1

AlphaNumeric

10

ax^{2}+bx+c = a\left( x^{2}+\frac{b}{a}x+\frac{c}{a}\right)

= a \left( x^{2} + 2\frac{b}{2a}x + \frac{c}{a}\right)

Unparseable latex formula:

= a \left( x^{2} + 2\frac{b}{2a}x + (\frac{b}{2a})^{2} - (\frac{b}{2}a})^{2} + \frac{c}{a}\right)

Unparseable latex formula:

= a \left( \left[ x^{2} + 2\frac{b}{2a}x + (\frac{b}{2a})^{2} \right] - (\frac{b}{2}a})^{2} + \frac{c}{a}\right)

= a \left( \left[x+\frac{b}{2a}\right]^{2}- (\frac{b}{2a})^{2} + \frac{c}{a} \right)

= a \left(x+\frac{b}{2a}\right)^{2} + c - \frac{b^{2}}{4a}

If a>0 then you can put the a back into the square root to give

ax^{2}+bx+c = \left( \sqrt{a}x + \frac{b}{2\sqrt{a}}\right)^{2} + c - \frac{b^{2}}{4a}

Consider

3x^{2}-8x-3=0

, a=3, b=-8 and c=-3, so putting this into

ax^{2}+bx+c = \left( \sqrt{a}x + \frac{b}{2\sqrt{a}}\right)^{2} + c - \frac{b^{2}}{4a}

gives

0=3x^{2}-8x-3 = \left( \sqrt{3}x + \frac{-8}{2\sqrt{3}}\right)^{2} + (-3) - \frac{(-8)^{2}}{4(3)}

\left( \sqrt{3}x + \frac{-8}{2\sqrt{3}}\right)^{2} = \frac{64}{12}+3

\left( \sqrt{3}x - \frac{4}{\sqrt{3}}\right)^{2} = \frac{100}{12}

\sqrt{3}x - \frac{4}{\sqrt{3}} = \pm \frac{5}{\sqrt{3}}

\sqrt{3}x = \frac{4}{\sqrt{3}} \pm \frac{5}{\sqrt{3}}

x = \frac{4}{3} \pm \frac{5}{3}

Therefore x = -1/3 and 3. Job is a good un'.

Reply 2

D.Hilbert

7

This would have been much easier, first try to see if this works, it sometimes does!:

3x^2 - 9x - 12 = 0

x^2 - 3x - 4 = 0

(x - 4)(x + 1) = 0

x = 4
x = -1

EDIT: Just as an example, doesn't work in every case I'm affraid (which would save you a LOT of time :frown:

)

Reply 3

fblade

14

Rickard.N

This would have been much easier, first try to see if this works, it sometimes does!:

3x^2 - 9x - 12 = 0

x^2 - 3x - 4 = 0

(x - 4)(x + 1) = 0

x = 4
x = -1

Thats wrong, the question is: 3x^2 - 8x- 3

Reply 4

D.Hilbert

7

I know the question, but he asked for a general answer as well. And I just suggested he'd try using the easier method first, since I said it sometimes works, and I gave an example...

Btw, the method aint wrong just because I didn't use it with his example (since it doesn't work in that case), it was just a tip :smile:

Reply 5

cuddy

3

the whole idea of completing the square is that you cant factorise it like you did so that example is pointless. Its not completing the square!

Reply 6

fblade

14

Rickard.N

I know the question, but he asked for a general answer as well. And I just suggested he'd try using the easier method first, since I said it sometimes works, and I gave an example...

Btw, the method aint wrong just because I didn't use it with his example (since it doesn't work in that case), it was just a tip :smile:

seeen my bad lol, yeh i get wht you mean :smile:

Anyone any good at compelting the sqauer i need help!!!

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