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# Easy C1 Question Watch

1. Find the values of k for which the equation x^2 - kx + (k+3) = 0 has real roots.

How would I go about doing this? Help would be much appreciated.
2. What is b^2 - 4ac (the discriminant) equal to when an equation has real roots?
3. you know the discriminant must be greater than zero.

D = b^2-4ac
4. real roots so . So . Just solve the inequality ( you do know how to, right?).

EDIT: if an equation has real roots, then . If an equation has equal roots, then . If an equation has no real roots, then

5. Of course. K is less than or equal to 2 and greater than or equal to 6. Thank you all

Edit: -2 not 2
6. (Original post by mya369)

Of course. K is less than or equal to 2 and greater than or equal to 6. Thank you all
Incorrect. It is less than 2 and greater than 6. By putting -2 in, you get: . For that equation, - so it has to be less than -2. If you put 6 into the equation, then it will be: . is equal to zero again. It is better to draw a curve going through both points and then look at the part above the x axis for >0 or the part below the x axis for <0.
7. (Original post by mya369)

Of course. K is less than or equal to 2 and greater than or equal to 6. Thank you all
actually i think it's -6 and 2

8. (Original post by MostCompetitive)
Incorrect. It is less than 2 and greater than 6. By putting -2 in, you get: . For that equation, - so it has to be less than -2. If you put 6 into the equation, then it will be: . is equal to zero again. It is always easier to draw a curve going through both points then looking at the part above the x axis for >0 or the part below the x axis for <0.
pretty sure "real roots" includes the situation of a repeated root. if it said "distinct roots", i'd agree with you
9. (Original post by MostCompetitive)
Incorrect. It is less than 2 and greater than 6. By putting -2 in, you get: . For that equation, - so it has to be less than -2. If you put 6 into the equation, then it will be: . is equal to zero again. It is better to draw a curve going through both points and then look at the part above the x axis for >0 or the part below the x axis for <0.
I think that 2 and 6 are inclusive. Repeated roots are still real roots.
10. (Original post by MostCompetitive)
Incorrect. It is less than 2 and greater than 6. By putting -2 in, you get: . For that equation, - so it has to be less than -2. If you put 6 into the equation, then it will be: . is equal to zero again. It is better to draw a curve going through both points and then look at the part above the x axis for >0 or the part below the x axis for <0.
Gunna go with equal roots being included. Thanks though
11. (Original post by Pheylan)
pretty sure "real roots" includes the situation of a repeated root. if it said "distinct roots", i'd agree with you
It doesn't. That is because when an equation has real roots, . That doesn't include equal to zero.
12. (Original post by mya369)
Gunna go with equal roots being included. Thanks though
> than 0 doesn't include equal to 0. Because i've done this paper recently, I can show you the mark scheme if you want.
13. (Original post by MostCompetitive)
> than 0 doesn't include equal to 0. Because i've done this paper recently, I can show you the mark scheme if you want.
I'm sure you can, but I found this question in an edexcel textbook. Their answer and the answers to similar questions include equal roots.

I agree with you in that if they state that for real roots the discriminant is greater than 0 (which they do), but not greater than or equal to 0, they shouldn't include equal roots in their answer. They aren't being consistent.
14. (Original post by mya369)
. They aren't being consistent.
I agree

it forms equal roots when k is 6, or when k is -2. There is no "less than.." or "greater than.." involved.

E.g when k = 6, it forms x^2 - 6x + 9 = (x - 3)(x - 3)

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