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# Find the range of values of k for which the equation x^2 - kx + (k+3) = 0 has no real Watch

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1. Find the range of values of k for which the equation x^2 - kx + (k+3) = 0 has no real roots.

Could I have some help with this question please?

I have got this far:

b^2 - 4ac < 0

k^2 - (4 x 1 x (k+3)) < 0

k^2 - 4k +12 = 0

(k - 6) (k + 2) = 0

k = 6 or k = -2

Afterwards I attempted to draw a graph of k^2 - 4k +12 = 0, however it does not cross the x-axis? Firstly, how is it possible to find solutions to an equation that does not cross the x-axis?

I was going to shade in the areas on the below the x-axis, as this would've given me values for where the equation is less than 0, but it doesn't cross the x-axis?

How should I proceed? Thanks for any help!
2. For no real roots
3. b^2 is negative rather than positive (check the original quadratic).
4. (Original post by TheOtherSide.)
b^2 is negative rather than positive (check the original quadratic).

b is -k

b^2 is (-k)^2, which is k^2

Am I missing something?
5. (Original post by NotNotBatman)
For no real roots

yeah sorry that was a silly slip when I was typing out my working, I shall edit it now, thanks for spotting it
6. (Original post by TheOtherSide.)
b^2 is negative rather than positive (check the original quadratic).
b is negative (-k), but b^2 is positive (-k)^2 = k^2
7. (Original post by tazarooni89)
b is negative (-k), but b^2 is positive (-k)^2 = k^2
That's what I meant - I probably should have been clearer..

(Original post by machete-monster1)

b is -k

b^2 is (-k)^2, which is k^2

Am I missing something?
Your actual equation would be -k^2 - 4k - 12, which suggests what about the shape of the curve?
8. (Original post by machete-monster1)
Find the range of values of k for which the equation x^2 - kx + (k+3) = 0 has no real roots.

Could I have some help with this question please?

I have got this far:

b^2 - 4ac < 0

k^2 - (4 x 1 x (k+3)) < 0

k^2 - 4k +12 = 0

(k - 6) (k + 2) = 0

k = 6 or k = -2

Afterwards I attempted to draw a graph of k^2 - 4k +12 = 0, however it does not cross the x-axis? Firstly, how is it possible to find solutions to an equation that does not cross the x-axis?

I was going to shade in the areas on the below the x-axis, as this would've given me values for where the equation is less than 0, but it doesn't cross the x-axis?

How should I proceed? Thanks for any help!
check that bolded line, you have made a mistake in expanding the bracket above

but you've somehow factorized right
9. (Original post by TheOtherSide.)
That's what I meant - I probably should have been clearer..

Your actual equation would be -k^2 - 4k - 12, which suggests what about the shape of the curve?
The equation is B^2 - 4AC < 0
Where B = (-k)
=> k^2 - 4(k+3) < 0

There should still be no negative sign before the k^2

The curve is a "U" shaped parabola, not an "n" shaped one.*
10. (Original post by tazarooni89)
The equation is B^2 - 4AC < 0
Where B = (-k)
=> k^2 - 4(k+3) < 0

There should still be no negative sign before the k^2

The curve is a "U" shaped parabola, not an "n" shaped one.*
Okay, I understand. But surely then the second equation will have no solutions?
11. (Original post by TheOtherSide.)
Okay, I understand. But surely then the second equation will have no solutions?
Wait, I see.
12. OP, check the y-intercept in the second equation..
13. (Original post by TheOtherSide.)
OP, check the y-intercept in the second equation..

Y intercept is 12

Is this incorrect?
14. (Original post by machete-monster1)

Y intercept is 12

Is this incorrect?
Slightly - check your bracket expansion.
15. (Original post by machete-monster1)
Find the range of values of k for which the equation x^2 - kx + (k+3) = 0 has no real roots.

Could I have some help with this question please?

I have got this far:

b^2 - 4ac < 0

k^2 - (4 x 1 x (k+3)) < 0

k^2 - 4k +12 = 0

(k - 6) (k + 2) = 0

k = 6 or k = -2

Afterwards I attempted to draw a graph of k^2 - 4k + 12 = 0, however it does not cross the x-axis? Firstly, how is it possible to find solutions to an equation that does not cross the x-axis?

I was going to shade in the areas on the below the x-axis, as this would've given me values for where the equation is less than 0, but it doesn't cross the x-axis?

How should I proceed? Thanks for any help!
You used the wrong sign here (should be k^2 - 4k - 12). Also, you should keep the inequalities in the equation, then sketch the graph to find your values of k.
16. Thank you very much everyone for the help; this was such a silly mistake!

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