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1. the question is:

f(x) = x2 − 5x + 10
Hence, or otherwise, determine whether f(x + 2) − 3 = 0 has any real roots

I know how to dertermine if it has any real roots, but how do i work out the 'f(x + 2) − 3' part?
2. I'm not too sure about this~ I haven't seen anything like it on Edexcel GCSE, but I'll try qwq

You should substitute x+2 into f(x) for every x value (for example, 3x would become 3(x+2)); you'd then expand the 2 brackets made and substitute your results into f(x+2)-3=0. Using a graphing calculator may help you if you're a bit lost on this (simplifying shouldn't be an issue), and the result for x goes into imaginary numbers.

...I think o3o I guess that'd prove that it has no real roots. I hope that I am right/have helped >.<
3. 1) Substitute x for (x+2) in f(x). You would end up with (x+2)^2 - 5(x+2) + 10 = 0
2) Then subtract 3 to get (x+2)^2 - 5(x+2) + 7 = 0
3) Expand brackets to get x^2 - x + 21 = 0
4) Use the discriminant b^2-4ac like you would with a quadratic
a = 1
b= -1
c= 21
For the discriminant:
If b^2 - 4ac is less than zero than the quadratic has no real roots
If b^2 - 4ac is equal to zero then the quadratic has two real and equal roots
If b^2 - 4ac is greater than zero than the quadratic has two real and different roots

So for this quadratic:
b^2 - 4ac = (-1)^2 - (4 x 1 x 2) = -7
Therefore b^2-4ac is less than zero, so the quadratic has no real roots

It's quite a hard question but if you break it down into stages it's actually several moderately easy topics combined into one daunting question
4. (Original post by maxtaylor16)
1) Substitute x for (x+2) in f(x). You would end up with (x+2)^2 - 5(x+2) + 10 = 0
2) Then subtract 3 to get (x+2)^2 - 5(x+2) + 7 = 0
3) Expand brackets to get x^2 - x + 21 = 0
4) Use the discriminant b^2-4ac like you would with a quadratic
a = 1
b= -1
c= 21
For the discriminant:
If b^2 - 4ac is less than zero than the quadratic has no real roots
If b^2 - 4ac is equal to zero then the quadratic has two real and equal roots
If b^2 - 4ac is greater than zero than the quadratic has two real and different roots

So for this quadratic:
b^2 - 4ac = (-1)^2 - (4 x 1 x 2) = -7
Therefore b^2-4ac is less than zero, so the quadratic has no real roots

It's quite a hard question but if you break it down into stages it's actually several moderately easy topics combined into one daunting question
Sorry, but I don't see how c is 21, nor how you changed it into 2 for the second bit (my calculator got 1 for c? 4-10+7). Thanks for explaining the rest though :3
5. (Original post by Latimaster)
Sorry, but I don't see how c is 21, nor how you changed it into 2 for the second bit (my calculator got 1 for c? 4-10+7). Thanks for explaining the rest though :3
Sorry! I got plus 10 because i wrote it down as (x-2) instead of (x+2).
6. Ah, that's okay then :3
7. You don't need to expand here. Just consider minimum point of f(x) and then consider whether moving it down 3 units would affect the number of real roots.

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