# Need Help with Surds Questions

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Given that the point with coordinates (1+sqrt3 , 5sqrt3) lies on the curve y=2x^2+px+q

Find the values of the rational constants p and q

And also this one, simplify:

(sqrt p+1 - sqrt p)(p+0.5+sqrt p^2+p) all over sqrt p+1 +sqrt p

Answers are: p=1 q=9 and the simplify is 1/2

I’m aware these are fairly simple questions, but we’ve moved on from the surds topic now and everything else was fairly comprehendable but I am unsure how to answer these 2 after doing the first simple parts.

Find the values of the rational constants p and q

And also this one, simplify:

(sqrt p+1 - sqrt p)(p+0.5+sqrt p^2+p) all over sqrt p+1 +sqrt p

Answers are: p=1 q=9 and the simplify is 1/2

I’m aware these are fairly simple questions, but we’ve moved on from the surds topic now and everything else was fairly comprehendable but I am unsure how to answer these 2 after doing the first simple parts.

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#2

(Original post by

Given that the point with coordinates (1+sqrt3 , 5sqrt3) lies on the curve y=2x^2+px+q

Find the values of the rational constants p and q

And also this one, simplify:

(sqrt p+1 - sqrt p)(p+0.5+sqrt p^2+p) all over sqrt p+1 +sqrt p

Answers are: p=1 q=9 and the simplify is 1/2

I’m aware these are fairly simple questions, but we’ve moved on from the surds topic now and everything else was fairly comprehendable but I am unsure how to answer these 2 after doing the first simple parts.

**Jason03**)Given that the point with coordinates (1+sqrt3 , 5sqrt3) lies on the curve y=2x^2+px+q

Find the values of the rational constants p and q

And also this one, simplify:

(sqrt p+1 - sqrt p)(p+0.5+sqrt p^2+p) all over sqrt p+1 +sqrt p

Answers are: p=1 q=9 and the simplify is 1/2

I’m aware these are fairly simple questions, but we’ve moved on from the surds topic now and everything else was fairly comprehendable but I am unsure how to answer these 2 after doing the first simple parts.

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(Original post by

What do you get when you sub the point into the first equation?

**mqb2766**)What do you get when you sub the point into the first equation?

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#4

(Original post by

5sqrt3 = 8+4sqrt3+p+psqrt3 +q

**Jason03**)5sqrt3 = 8+4sqrt3+p+psqrt3 +q

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(Original post by

So choose p & q (rational) to make the surds and rational parts match on each sude.

**mqb2766**)So choose p & q (rational) to make the surds and rational parts match on each sude.

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#6

(Original post by

Not too sure what you mean, just answer it by trial and error?

**Jason03**)Not too sure what you mean, just answer it by trial and error?

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(Original post by

No, for the surds, what multiplies sqrt(3) on each side of the equation. They must balance.

**mqb2766**)No, for the surds, what multiplies sqrt(3) on each side of the equation. They must balance.

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#8

(Original post by

Yeah so it has to be p=1 and therefore q=9 but there isn’t any like law or anything I need to know, just that the 5sqrt3=4sqrt3+psqrt3 and then 0=8+p+q it’s that simple?

**Jason03**)Yeah so it has to be p=1 and therefore q=9 but there isn’t any like law or anything I need to know, just that the 5sqrt3=4sqrt3+psqrt3 and then 0=8+p+q it’s that simple?

Can you upload a pic of the second question, the layout is confusing.

Last edited by mqb2766; 4 weeks ago

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**mqb2766**)

Yes it's that simple. It just used basic arithmetic laws on rational and irrational (surds) numbers.

Can you upload a pic of the second question, the layout is confusing.

Last edited by Jason03; 4 weeks ago

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#10

(Original post by

**Jason03**)Look at the first term in the numerator and the complete denominator.

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(Original post by

What do you think you can do to rationalise it?

Look at the first term in the numerator and the complete denominator.

**mqb2766**)What do you think you can do to rationalise it?

Look at the first term in the numerator and the complete denominator.

then to 2p+1/2+sqrt p

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#13

Last edited by mqb2766; 4 weeks ago

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(Original post by

It s also with thinking how you can represent the second term in the numerator in terms of sqrt(p) terms. That's probably the most elegant way to solve the problem.

**mqb2766**)It s also with thinking how you can represent the second term in the numerator in terms of sqrt(p) terms. That's probably the most elegant way to solve the problem.

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#15

For the 2nd term on numerator

((p + 1/2 + sqrt(p^2+p))

1/2(2p + 1 + 2sqrt(p)sqrt(p+1))

1/2(p+1 + 2sqrt(p)sqrt(p+1) + p)

1/2(sqrt(p+1) + sqrt(p))^2

Easy to finish?

((p + 1/2 + sqrt(p^2+p))

1/2(2p + 1 + 2sqrt(p)sqrt(p+1))

1/2(p+1 + 2sqrt(p)sqrt(p+1) + p)

1/2(sqrt(p+1) + sqrt(p))^2

Easy to finish?

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(Original post by

For the 2nd term on numerator

((p + 1/2 + sqrt(p^2+p))

1/2(2p + 1 + 2sqrt(p)sqrt(p+1))

1/2(p+1 + 2sqrt(p)sqrt(p+1) + p)

1/2(sqrt(p+1) + sqrt(p))^2

Easy to finish?

**mqb2766**)For the 2nd term on numerator

((p + 1/2 + sqrt(p^2+p))

1/2(2p + 1 + 2sqrt(p)sqrt(p+1))

1/2(p+1 + 2sqrt(p)sqrt(p+1) + p)

1/2(sqrt(p+1) + sqrt(p))^2

Easy to finish?

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#17

(Original post by

Not sure how sqrt(p^2+p) becomes sqrt(p)sqrt(p+1) can you explain? Sorry for being a pest law

**Jason03**)Not sure how sqrt(p^2+p) becomes sqrt(p)sqrt(p+1) can you explain? Sorry for being a pest law

sqrt(ab) = sqrt(a)sqrt(b)

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#19

(Original post by

Yeah but surely wouldn’t it go to sqrt(p^2)sqrt(p) which would just become (p)sqrt(p) or am I missing something

**Jason03**)Yeah but surely wouldn’t it go to sqrt(p^2)sqrt(p) which would just become (p)sqrt(p) or am I missing something

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#20

**Jason03**)

Yeah but surely wouldn’t it go to sqrt(p^2)sqrt(p) which would just become (p)sqrt(p) or am I missing something

Then root.

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