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# Maths GCSE Exam watch

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1. 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
2. (Original post by 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?
3. (Original post by 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.
4. 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
5. (Original post by samdavyson)
Ok.

Is that true? Does no one give you rep?
well yes nobody gives it me! maybe that my contributions are worthless!
6. (Original post by 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!!!)
7. (Original post by 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!
8. (Original post by 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!
9. (Original post by 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

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Updated: May 31, 2004
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