C1 help

Fraggr

Fraggr

Fraggr

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Joined
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Sheffield
For the past 2 days, I've been doing M1 revision, and I've got pretty good at it (96% on the timed practice paper today). So now I have turned my attention to C1 and C2.

The stuff in C1 I can do pretty easily normally. But there's a question I am doing at the moment, and I am stuck on it.

5. Solve the simultaneous equations
x – 2y = 1,
x2 + y2 = 29.

I have got x = 5 when y = 2. Easy. But now I have to substitute y = -14/5 into the quadratic, and I see no way of doing it without a calculator... I know the answer is x = -23/5 when y = -14/5 (mark scheme is on the website I got the test from), but can't see how to get it without using a calculator for the square roots...

I managed to get (14/5) squared = 7.84 by writing it out, but can't see how to do sqrt(29-7.84) in my head/paper...

Any help?
 
Rearrange the first equation in terms of one variable (x or y, your choice really - x is probably easier because it doesn't have a coefficient in front of it) and then substitute that value into the first to give you a numerical value for the other variable. Once you know that value you can put it back into the first equation to work out the numerican value of the other.
 
I know that stuff. It's how I worked out the first part. But how do I do stuff like finding the square root of something like 21.16 in my head..?
 
If you keep it in terms of fractions it's OK - you have (29*25)/25 - (14*14)/25 on the RHS, which simplifies down to 529 / 25. Although whether you're expected to recognise that 529 is the square root of 23 is another thing!
 
Dave said:
If you keep it in terms of fractions it's OK - you have (29*25)/25 - (14*14)/25 on the RHS, which simplifies down to 529 / 25. Although whether you're expected to recognise that 529 is the square root of 23 is another thing!
Click to expand...

Yeah. I had that.. I marked it off as being wrong for that reason (didn't notice that it was the square root). Thanks for that. :)
 
Fraggr said:
I know that stuff. It's how I worked out the first part. But how do I do stuff like finding the square root of something like 21.16 in my head..?
Click to expand...

You don't need to, here is how i would do it:
substitute the first equation into the second and rearrange to give the quadratic
(5y2 + 4y -28) = 0
you can then factorise that to get y=2 and y = -14/5

To find x you then substitute these values into the intial equations (either)

so use x-2y = 1 with y=-14/5
which rearranges to give the answer

Does that explain it a bit better or have i missed something?
 
dave_a141 said:
You don't need to, here is how i would do it:
substitute the first equation into the second and rearrange to give the quadratic
(5y2 + 4y -28) = 0
you can then factorise that to get y=2 and y = -14/5

To find x you then substitute these values into the intial equations (either)

so use x-2y = 1 with y=-14/5
which rearranges to give the answer

Does that explain it a bit better or have i missed something?
Click to expand...

I already had that. :)

I was asking for how to solve the equation where y = -14/5 is subbed in. It was a matter of noticing 23 is the square root of 529.
 
That is one way of doing it but its got to be easier to substitute y=-14/5 into the equation x-2y=1 instead of into the quadratic - then you get a much simpler rearrangement with no square roots?
 
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