# Thread: er quadratic fatorisation?

1. ## er quadratic fatorisation?

Any help on this would be appreciated...

write $\displaystyle 4x^2 - 24x + 27$ in the form $\displaystyle a(x-b)^2+c$

i have thought about factorising but it wants the $\displaystyle + c$ which just confuses me

thanks

2. Originally Posted by RubbishAtMaths
Any help on this would be appreciated...

write $\displaystyle 4x^2 - 24x + 27$ in the form $\displaystyle a(x-b)^2+c$

i have thought about factorising but it wants the $\displaystyle + c$ which just confuses me

thanks

$\displaystyle y=4x^2-24x+27$

$\displaystyle y-27=4x^2-24x$

$\displaystyle y-27=4(x^2-6x)$

$\displaystyle y-27=4(x^2-6x+9-9)$

$\displaystyle y-27=4(x-3)^2-4(9)$

$\displaystyle y=4(x-3)^2-9$

Edit: Moo beat me to it!

3. Hello,
Originally Posted by RubbishAtMaths
Any help on this would be appreciated...

write $\displaystyle 4x^2 - 24x + 27$ in the form $\displaystyle a(x-b)^2+c$

i have thought about factorising but it wants the $\displaystyle + c$ which just confuses me

thanks
$\displaystyle 4x^2-24x+27=4(x^2-6x)+27=4[(x-3)^2-3^2]+27$
$\displaystyle =4(x-3)^2-4 \cdot 3^2+27=4(x-3)^2-9$

4. how do you go about completing the square?

*thanks moo!

5. Originally Posted by RubbishAtMaths
how do you go about completing the square?

*thanks moo!
\

6. ## could u just explain this a little please

i understand most of it just not the bit i have pointed out
Originally Posted by TheMasterMind
$\displaystyle y=4x^2-24x+27$

$\displaystyle y-27=4x^2-24x$

$\displaystyle y-27=4(x^2-6x)$

$\displaystyle y-27=4(x^2-6x+9-9)$

$\displaystyle y-27=4(x-3)^2-4(9)$
I dont Understand what u did with the -27 and how u got the -4(9)???
thanks
$\displaystyle y=4(x-3)^2-9$

Edit: Moo beat me to it!

7. Originally Posted by adeshSB
i understand most of it just not the bit i have pointed out
$\displaystyle y-27=4(x-3)^2-9(4)$

$\displaystyle y-27=4(x-3)^2-36$

$\displaystyle y=4(x-3)^2-36+27$

$\displaystyle y=4(x-3)^2-9$

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