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Maths GCSE Exam

ax²+bx+c = a(x²+bx/a+c/a) = a[(x+b/2a)² - b²/4a² + c/a] = a[(x+b/2a)² - (4ac-b²)/4a²]

If you put this equal to 0 then you can go on to derive the quadratic formula

Reply 41

samd

10

gemma.....

oh, i see..umm but wasn't that orignial post just an example of Ax where the letter is the number? as in C squared = A squared, but the letter isn't a letter it represents a number?

Ignore me i don't understand lol

Ok.

Is that true? Does no one give you rep?

Reply 42

samd

10

Bezza

ax²+bx+c = a(x²+bx/a+c/a) = a[(x+b/2a)² - b²/4a² + c/a] = a[(x+b/2a)² - (4ac-b²)/4a²]

If you put this equal to 0 then you can go on to derive the quadratic formula

Thank you kindly.

Reply 43

m:)ckel

14

just to expand on what bezza said (only read it if you're curious....there's no need for you to know it for gcse):

ax² + bx + c = 0

a[x² + (b/a)x + (c/a)] = 0

(x + b/2a)² - (b²/4a²) + (c/a) = 0

(x + b/2a)² = (b²/4a²) - (c/a)

(x + b/2a)² = (b²-4ac) / 4a²

x + b/2a = ±√(b²-4ac) / 2a

x = -(b/2a) ±√(b²-4ac) / 2a

x = [-b±√(b²-4ac)] / 2a

Reply 44

G.A.K

4

samdavyson

Ok.

Is that true? Does no one give you rep?

well yes nobody gives it me! maybe that my contributions are worthless!

Reply 45

m:)ckel

14

gemma.....

well yes nobody gives it me! maybe that my contributions are worthless!

don't worry....keep contributing!
incidentally, i didn't get any rep for the first 10 months i was on here! now i've received it 7 times in the last two months (i realise this isn't much, but compared to no rep in 10 months!!!)

Reply 46

G.A.K

4

mockel

don't worry....keep contributing!
incidentally, i didn't get any rep for the first 10 months i was on here! now i've received it 7 times in the last two months (i realise this isn't much, but compared to no rep in 10 months!!!)

HaHa - well i did have a look and i did get some rep today without knowing - but seriously it doesn't matter, it owuld if i had been on here for ten months tho!

Reply 47

john !!

14

samdavyson

Could you run over that?

Is it just:

ax² + bx + c = 0

a(x + ½bx)² ... in fact I have no idea. Please enlighten.

ax² + bx + c = 0 --- divide everything by a to 'clean up' the x²
x² + (b/a)x + (c/a) = 0 --- first stange of completing the square (note 4ac/4a² = c/a)
(x + (b/2a))² - (b²/4a²) + (4ac/4a²) = 0 --- neaten up a bit
(x + (b/2a))² - ((b² - 4ac)/4a²) = 0 --- take the negative across and square root both sides
x + (b/2a) = (±√(b² - 4ac))/2a --- take (b/2a) from both sides, finishing with
x = (-b±√(b² - 4ac))/2a

et voila!

Reply 48

john !!

14

gemma.....

oh, i see..umm but wasn't that orignial post just an example of Ax where the letter is the number? as in C squared = A squared, but the letter isn't a letter it represents a number?

Ignore me i don't understand lol

A quadratic equation has the formula:

ax² + bx + c = 0

If we say there is a curve:
y = ax² + bx + c

Then we are describing how x and y are linked, and this can be plotted on a graph. For any particular curve, a, b, and c are constants - they do not change. They define the curve. For example, if you changed a from 2 to 4, you would have a steeper curve, because y is more 'sensetive' to changes in x.

When y = 0, the equations y = ax² + bx + c and ax² + bx + c = 0 mean the same thing. When y = 0, also, this line describes a horizontal line - the x-axis. So by putting y equal to zero we can find the x values of the curve where it crosses this axis.

For example,

4x² + 2x + 3 = 0 ------ (a=4, b=2, c=3)
x² - 2x + 7 = 0 ------- (a=1, b=-2, c=3)
9x² - 4 = 0 ----------- (a=9, b=0, c=-4)

To solve the last one as an example, we have to factorise by putting into brackets:

9x² - 4 = (3x + 2)(3x - 2) ------- (try multiplying out and checking this)

if we know that:

9x² - 4 = 0,
(3x + 2)(3x - 2) = 0,

Now since these two multiply to make zero, one of them must be zero. So we put each of the brackets equal to zero to find both solutions for x:

3x + 2 = 0
3x = -2
x = -2/3

3x - 2 = 0
3x = 2
x = 2/3

Maths GCSE Exam

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