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How do I work out what x is?

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Reply 20
Avatar for Kash:)
Kash:)
OP
15
TheNack
Just rearrange that quadratic equation you have there to get x2x^2 on one side, square root everything, and remember that you can have a positive or negative answer when you take a square root.

Sorry to be a pain and everything...

I've done

2x2=32x^2=3
x2=32x^2=\frac{3}{2}
x=±32x=\pm\sqrt\frac{3}{2}

I've checked with a calculator and it's correct.

But in my book it says the answer is ±2\pm\sqrt2 (the answers in my book have been wrong before though, if that is the case this time..)
Reply 21
Avatar for Kash:)
Kash:)
OP
15
Pheylan
How old are you?

A bit of an odd question... I am 18. Why?
Reply 22
Avatar for Pheylan
Pheylan
Kash:)
A bit of an odd question

lol sorry

are you doing A2 maths?
Reply 23
Avatar for Kash:)
Kash:)
OP
15
Pheylan
lol sorry

are you doing A2 maths?

AS. C1
Reply 24
Avatar for Pheylan
Pheylan
Kash:)
AS. C1

okey doke :smile:
Reply 25
Avatar for Kash:)
Kash:)
OP
15
Hi, I have some other questions I am having problems with now...

How would I find x for these?

y=2x3+x2y=2x^3+x^2

I've done this so far, really unsure if it is correct, and I don't know what to do next
0=2x3+x2 0=2x^3+x^2
2x3=−x2 2x^3=- x^2
2x2=−x 2x^2=-x


Another one I have is
y=x3−3x2+4 y=x^3-3x^2+4

I know I should factorise it, i've had a go but can't work out how to do it... :o:
Reply 26
Avatar for caffrey
caffrey
y=2x^3+x^2 => y=x^2(2x+1)
When y=0 (i.e. the curve crosses the x-axis), either:
i) x^2=0 => x=0
ii) 2x+1=0 => 2x=-1 => x=-0.5

y=x^3-3x^2+4 => y=(x+1)(x-2)^2 => Two solutions: x=-1, x=2
Reply 27
Avatar for Kash:)
Kash:)
OP
15
caffrey
y=2x3+x2=>y=x2(2x+1)y=2x^3+x^2 => y=x^2(2x+1)
When y=0 (i.e. the curve crosses the x-axis), either:
i)x2=0=>x=0 x^2=0 => x=0
ii) 2x+1=0=>2x=−1=>x=−0.52x+1=0 => 2x=-1 => x=-0.5

y=x3−3x2+4=>y=(x+1)(x−2)2=>y=x^3-3x^2+4 => y=(x+1)(x-2)^2 => Two solutions: x=-1, x=2



Thanks, I see it now. Think I just need to go do some practice on how to factorise things
Reply 28
Avatar for Kash:)
Kash:)
OP
15
caffrey
y=2x^3+x^2 => y=x^2(2x+1)
When y=0 (i.e. the curve crosses the x-axis), either:
i) x^2=0 => x=0
ii) 2x+1=0 => 2x=-1 => x=-0.5

y=x^3-3x^2+4 => y=(x+1)(x-2)^2 => Two solutions: x=-1, x=2

How did you manage to work out this?
Reply 29
Avatar for caffrey
caffrey
Kash:)
How did you manage to work out this?


To be honest I just stared at it for a while. If it was really difficult I might do trial and error, testing likely factors such as x+1. The factors aren't usually large numbers.

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