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# Core 1 quadratic function and inequalities watch

1. find the values of k for which the equation x2 - 2(k+1)x + 2k2 -7 = 0 , has real roots

anyone help on this question? I'm guessing I will have to use b2 - 4ac for this with b2 - 4ac > 0 as there are real roots?
Just thrown off a bit by having 2 x2 k2
values.
Thank you😊
2. Just ignore the X and use -2(K+1)
3. (Original post by Flask)
find the values of k for which the equation x2 - 2(k+1)x + 2k2 -7 = 0 , has real roots

anyone help on this question? I'm guessing I will have to use b2 - 4ac for this with b2 - 4ac > 0 as there are real roots?
Just thrown off a bit by having 2 x2 k2
values.
Thank you😊
Yes you use b^2-4ac > 0

A= 1
B= -2k-2
C= -7

Now just use the formula, where you will end up with a with a quadratic inequality in k, and then solve.

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4. (Original post by Jimmy20002012)
Yes you use b^2-4ac > 0

A= 1
B= -2k-2
C= -7

Now just use the formula, where you will end up with a with a quadratic inequality in k, and then solve.

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C is 2k2-7

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5. (Original post by brother aldi)
C is 2k2-7

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Haha misread it, my fault.

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6. (Original post by Flask)
find the values of k for which the equation x2 - 2(k+1)x + 2k2 -7 = 0 , has real roots

anyone help on this question? I'm guessing I will have to use b2 - 4ac for this with b2 - 4ac > 0 as there are real roots?
Just thrown off a bit by having 2 x2 k2
values.
Thank you😊

7. Worked through: - (Where '^2' = Squared)
Using b^2-4ac:

a=1
b=-2k-2
c=2k^2-7

b^2-4ac = (-2k-2)^2 -4(2k^2-7)>0
=(4k^2+8k+4) -4(2k^2-7)>0
=4k^2+8k+32 -8k^2+28 >0
=4k^2-8k-32<0
=k^2-2k-8<0=(k-4)(k+2)<0
= -2<k<4
8. (Original post by AndrewC19)
Worked through: - (Where '^2' = Squared)
Using b^2-4ac:

a=1
b=-2k-2
c=2k^2-7

b^2-4ac = (-2k-2)^2 -4(2k^2-7)>0
=(4k^2+8k+32) -4(2k^2-7)>0
=4k^2+8k+32 -8k^2+28 >0
=4k^2-8k-32<0
=k^2-2k-8<0=(k-4)(k+2)<0
=k<-2 k>4 (I may have this last part the wrong way round).
This is wrong. You shouldn't post full solutions, let OP work through the question themselves.
9. (Original post by lizard54142)
This is wrong. You shouldn't post full solutions, let OP work through the question themselves.
I will delete it, I didn't realise. New to this, sorry.
10. (Original post by AndrewC19)
I will delete it, I didn't realise. New to this, sorry.
No problem it's just much better for people to work through a question with only a few pointers rather than just tell them the right answer. Also not sure how you got (-2)^2 = 32 in line 1, and check the inequalities last line!
11. (Original post by lizard54142)
No problem it's just much better for people to work through a question with only a few pointers rather than just tell them the right answer. Also not sure how you got (-2)^2 = 32 in line 1, and check the inequalities last line!
Lol you can post a worked question if you like, don't listen to this fool
12. That was a mistake, I did this on paper first you see. I meant to put +4
I get confused with the inequalities, I wasn't ever taught it. I just guess.

Could you tell me how to know when to use which inequality? (If that makes sense )
13. (Original post by shamsaidk)
Lol you can post a worked question if you like, don't listen to this fool

14. (Original post by AndrewC19)
That was a mistake, I did this on paper first you see. I meant to put +4
I get confused with the inequalities, I wasn't ever taught it. I just guess.

Could you tell me how to know when to use which inequality? (If that makes sense )
were you taught how to draw the graphs associated with the question?
15. Briefly. Should it have been -2<k<4?
16. (Original post by AndrewC19)
That was a mistake, I did this on paper first you see. I meant to put +4
I get confused with the inequalities, I wasn't ever taught it. I just guess.

Could you tell me how to know when to use which inequality? (If that makes sense )
Consider the graph of y = (x-4)(x+2).

For what values of x is y < 0 ?
17. (Original post by AndrewC19)
Briefly. Should it have been -2<k<4?
Yep
18. Ok this makes sense now. It's coming back to me!
Thanks so much Lizard
19. (Original post by AndrewC19)
Ok this makes sense now. It's coming back to me!
Thanks so much Lizard
No problem as soon as you see an inequality, think graph!
20. (Original post by lizard54142)
No problem as soon as you see an inequality, think graph!
I'll keep this fresh in my head ready for Wednesday. Thanks again.

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