James Dale

Posted on Sunday, 30 May, 2004 - 12:29 pm:

when using the formula

$x = (- b \pm \sqrt{b^{2} - 4 a c}) / 2 a$

i have no problem solving equations like

$2 x^{2} - 5 x - 4 = 0$

but i do not know how to solve an equation such as:

$x^{2} - 5 x + 6 = 0$

because we usually deal with $a x^{2} + b x + c = 0$ where $a$ , $b$ and $c$ are numbers but with $x^{2}$ you are not given a number for $a$ so i cant solve the equation. if it is $x^{2}$ does the number become 1? please help guys cheers

Michael McLoughlin

Posted on Sunday, 30 May, 2004 - 12:48 pm:

If you wish to use the formula for the equation x² -5x+6=0 you take a=1, b=-5, c=6. However, this particular equation can be solved in a much easier way since x² -5x+6 = (x-2)(x-3) = 0. Therefore, x-2=0 or x-3=0 which means that the solutions are x=2 or x=3.

Matthew Buckley

Posted on Sunday, 30 May, 2004 - 02:42 pm:

James, you just think of x² as 1*x² (because multiplying anything by 1 doesn't change it's value), so a = 1 in the formula in this instance.

Matt.

Nicola Coles

Posted on Sunday, 30 May, 2004 - 02:47 pm:

x² = 1 times x² = 1x²

So x² -5x+6 = 1x² -5x+6

So yes, a=1 in this case.

Waqass Farid

Posted on Sunday, 20 June, 2004 - 11:06 pm:

THERE IS MORE THAN ONE WAY OF SOLVING A QUADRATIC EQUATION.
ALTHOUGH YOUR METHOD IS CORRECT IT IS OFTEN EASIER TO USE "COMPLETING THE SQUARE" METHOD WHEN PRESENTED WITH NO QUANTITY BEFORE X SQUARED.
YOU CANNOT ALWAYS USE MICHAELS METHOD WITH EVERY QUADRATIC EQUATION.

WAQ

Matthew Buckley

Posted on Monday, 21 June, 2004 - 12:14 am:

Well technically you can....... you'd just need a heck of a lot of insight generally as to what the factors were though!

Sorry - just me being pedantic!

Matt.

Andrew Fisher

Posted on Monday, 21 June, 2004 - 12:29 am:

Perhaps worth pointing out that using the formula originally given is effectively the same as completeing the square.