MATHS C1: What&#x27;s -b/2a?

Original post by creativebuzz

I got the question right (x = -4), however the mark scheme's example used -b/2a. I know that -b/2a is a quick method of finding the vertex of a graph, but I don't understand how that's relevant to this question.

(I'm referring to part (B))

since the quadratic is a perfect square x=-b(2a) is the solution

Reply 3

OP

Original post by TeeEm

since the quadratic is a perfect square x=-b(2a) is the solution

Can you explain that further? (sorry, I just really want to understand this completely)

Reply 4

OP

Original post by TeeEm

Line of symmetry of a quadratic is -b/(2a) following the usual notation

Yeah I know that it's the vertex etc! I just don't understand how that relates to the question..

Reply 5

Original post by creativebuzz

Can you explain that further? (sorry, I just really want to understand this completely)

if p=4

you can sub in and get x² - 8x +16 = 0
perfect square expected as (x-4)²
solution x = 4 (repeated)

or

you do not bother

x=-b/(2a)=2p/(2x1) = 4 again

Reply 6

ztibor

10

Original post by creativebuzz

I got the question right (x = -4), however the mark scheme's example used -b/2a. I know that -b/2a is a quick method of finding the vertex of a graph, but I don't understand how that's relevant to this question.

(I'm referring to part (B))

This come from the following quadratic eq. formula

\displaystyle x_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}

For 2 equal solutions the

D=\sqrt{b^2-4ac}

(discriminant) is zero
From this you will get the value of p
where (from the equation)
a=1
b=2p
c=3p+4

So the
solution of the equation with this value of p will be

\displaystyle x_{1,1}=\frac{-b \pm D}{2a}=\frac{-b+0}{2a}=-\frac{b}{2a}

Reply 7

Original post by creativebuzz

Can you explain that further? (sorry, I just really want to understand this completely)

eqaul roots =perfect square
u can prove this results with the 'general' roots

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Reply 8

Original post by physicsmaths

eqaul roots =perfect square
u can prove this results with the 'general' roots

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i meant prove the -b/2a result proof

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Reply 9

OP

Original post by physicsmaths

eqaul roots =perfect square
u can prove this results with the 'general' roots

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I don't understand what you're trying to say..

Reply 10

OP

Original post by TeeEm

if p=4

you can sub in and get x² - 8x +16 = 0
perfect square expected as (x-4)²
solution x = 4 (repeated)

or

you do not bother

x=-b/(2a)=2p/(2x1) = 4 again

Oh okay! So what has that got to do with the vertex?

Reply 11

interstitial

by symmetry, the line of symmetry of a quadratic is halfway between the roots (ie line of symmetry is

x=\dfrac{\alpha+\beta}{2}

where alpha and beta are the roots of the equation)

it just so happens that the formula for

\alpha+\beta

is -b/a (if you do Fp1, you'll find out more about this), so the line of symmetry must be at -b/2a

(edited 9 years ago)

Reply 12

consider

x= -b+-sqrt(B^2-4ac)/2a
since b^2-4ac =0
this makes x=-b/2a

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Reply 13

Original post by creativebuzz

Oh okay! So what has that got to do with the vertex?

the vertex is a touching point on the x axis! :smile:

Reply 14

OP

Original post by Arithmeticae

by symmetry, the x coordinate vertex of a quadratic is halfway between the roots (e.g.

\dfrac{\alpha+\beta}{2}

)

it just so happens that the formula for

\alpha+\beta

is -b/a (if you do Fp1, you'll find out more about this), so the x coordinate of the vertex (and hence the line of symmetry) must be at -b/2a

Oh I see now! So will this work with any coordinate or do the number have to be a perfect square?

Reply 15

OP

Original post by TeeEm

the vertex is a touching point on the x axis! :smile:

Ah of course, because it's a repeated root! Thank you!

So if the question didn't include 4/aka a perfect square, for example if it was three (a number which isn't a perfect square), could I still use the formula -b/2a or?

Reply 16

Original post by creativebuzz

Ah of course, because it's a repeated root! Thank you!

So if the question didn't include 4/aka a perfect square, for example if it was three (a number which isn't a perfect square), could I still use the formula -b/2a or?

look at the statement i made above. u can use it wheneve b^2-4ac=0

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Reply 17

interstitial

Original post by creativebuzz

Oh I see now! So will this work with any coordinate or do the number have to be a perfect square?

the line of symmetry (and hence the x coordinate of teh vertex) will always be at -b/2a

Reply 18

Original post by creativebuzz

Ah of course, because it's a repeated root! Thank you!

So if the question didn't include 4/aka a perfect square, for example if it was three (a number which isn't a perfect square), could I still use the formula -b/2a or?

by perfect square they dont mean a number the mean it is factorised into a dohble root ie
(x-1)^2 or
(x-33)^2

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Reply 19