Core 1 quadratic function and inequalities

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 x²k²
values.
Thank you😊

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Reply 1

SanjayGill

Just ignore the X and use -2(K+1)

Reply 2

Jimmy20002012

Original post by Flask

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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Reply 3

brother aldi

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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Reply 4

Jimmy20002012

Original post by brother aldi

C is 2k2-7

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Haha misread it, my fault. :smile:

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Reply 5

lizard54142

Original post by Flask

ax^2 + bx + c = 0

For your question:

a = 1

b = -2(k+1)

c = 2k^2 - 7

Reply 6

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+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

(edited 8 years ago)

Reply 7

lizard54142

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.

Reply 8

AndrewC19

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.

Reply 9

lizard54142

Original post by AndrewC19

I will delete it, I didn't realise. New to this, sorry.

No problem :smile:

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!

Reply 10

shamsaidk

Original post by lizard54142

No problem :smile:

Lol you can post a worked question if you like, don't listen to this fool

Reply 11

AndrewC19

That was a mistake, I did this on paper first you see. I meant to put +4 :smile:

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 :confused:

)

Reply 12

lizard54142

Original post by shamsaidk

Lol you can post a worked question if you like, don't listen to this fool

http://www.thestudentroom.co.uk/showthread.php?t=403989

Read the bit about full solutions.

Reply 13

shamsaidk

Original post by AndrewC19

That was a mistake, I did this on paper first you see. I meant to put +4 :smile:

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 :confused:

)

were you taught how to draw the graphs associated with the question?

Reply 14

AndrewC19

Briefly. Should it have been -2<k<4?

Reply 15

lizard54142

Original post by AndrewC19

That was a mistake, I did this on paper first you see. I meant to put +4 :smile:

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 :confused:

)

Consider the graph of y = (x-4)(x+2).

For what values of x is y < 0 ?