For what values of k will the equation have no real roots?

  • #51
SammyS
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3+k
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(3+k)^2-4(2)(1/2)
=(3+k)(3+k)-4
=9+3k+3k+k^2-4
=k^2+6k+5

b^2 - 4ac

(6k)^2-4(k^2)(5)
(6k)(6k)-20k^2
36k^2-20k^2
16k^2
Don't look at the discriminant for k2 + 6k + 5 = 0 .

Look again at what Sgt. Schultz jbriggs444 said in Post #46

...

You want values of k for which the discriminant is non-negative. You have identified values of k for which the discriminant is zero.

Edit: To be clear, you are on the right track and doing well now.
Edit: And reading back to the original problem statement we want values of k for which the discriminant is negative.
k2 + 6k + 5 is the discriminant for the original quadratic equation.

Solve the inequality: k2 + 6k + 5 < 0 .
 
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  • #52
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2
3+k
1/2

(3+k)^2-4(2)(1/2)
0=(3+k)(3+k)-4
0=9+3k+3k+k^2-4
0=k^2+6k+5
(k+1)(k+5)<0
-1>k>-5
 
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  • #53
MarcusAgrippa
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You are going in circles. Your original equation was
[itex] 2x^2-3x+kx=-1/2[/itex]
In standard form, the quadratic equation is
[itex] ax^2+bx+c = 0[/itex]
So your equation, in standard form becomes
[itex] 2x^2 +(k-3) x + 1/2 = 0[/itex]
or, multiplying by 2,
[itex] 4x^2 + 2(k-3) x + 1 = 0[/itex]
Comparing with the standard form, this gives
[itex] a= 4, b= 2(k-3) , c= 1 [/itex]
The discriminant of the equation in standard form is defined to be
[itex] b^2 - 4 a c [/itex]
The discriminant is the quantity under the square root sign in the general solution of the quadratic equation
[itex] x = \frac{-b \pm \sqrt{b^2 - 4 a c}}{2a} [/itex]
You get no real roots when the square root does not yield a real number, that is, when the quantity under the square root is less than zero.

So, in your case, what is the discriminant?

When you have answered, your equation has no real roots when the discriminant is less than zero. So you must put
[itex] b^2 - 4 a c < 0 [/itex]
Finally, you must solve the inequality.

Now try it for yourself. You must try to solve it yourself, otherwise the help you get is not help.

Waar woon jy, Jaco?
 
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  • #54
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Hi Marcus,
Yes, i follow.
 
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  • #55
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(2(k-3))^2-16<0
(2(k-3)(k-3)-16<0
4(k-3)^2-16<0
4((k-3)^2-4)<0
k^2-6K-4+9<0
k^2-6k+5<0
 
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  • #56
MarcusAgrippa
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Ok. Have you read my edited version of the last post?

Ok. Now we are getting somewhere. Jaco, why do you have a string of mathematical statements with no indication of how the statements are connected?

You should be able to read mathematics in the same way that you read English! It must make sense! Mathematics is a very concise shorthand. It is a language in its own right that abbreviates and compactifies language in order to make the application of logic to your statements transparent.
 
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  • #57
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(2(k-3))^2-16
(2(k-3)(k-3)-16
4(k-3)^2-16
4((k-3)^2-4)
k^2-6K-4+9
k^2-6k+5
Tell us what you are doing here. What you wrote here is the mathematical equivalent of talking to someone without using any verbs. Furthermore, you start off with this: (2(k-3)(k-3)-16. At least include some words to help us understand what you're doing. (I know you're working with the discriminant, but you should say that.)
Ok. Have you read my edited version of the last post?

Ok. Now we are getting somewhere. Jaco, why do you have a string of mathematical statements with no indication of how the statements are connected?
In fact, these aren't even statements (which would include equations or inequalities) -- they are just expressions. Even so, I understand what you're saying @MarcusAgrippa, and I agree completely. What Jaco has written is a bunch of expressions that might as well be random scribbling. There are no connections (such as = , >, or < ) in the whole batch.
 
  • #58
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Sorry guys I am getting lost in this, I feel like I am doing the same thing over and over.
to and fro (b^2−4ac)&(ax2+bx+c=0)

Its been a long time since Ive done math.

Marcus, Ek is in Alberton, en jy?
I see you have edited again.

Should I be using this:


I will try and go through some videos and see if I can get my head around it.
 
  • #59
SammyS
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(3+k)^2-4(2)(1/2)
0=(3+k)(3+k)-4
0=9+3k+3k+k^2-4
0=k^2+6k+5
(k+1)(k+5)<0

-1>k>-5
That is the correct final result for k.

The working of this leaves a gap from
0=k2+6k+5​
to
(k+1)(k+5)<0​

You go from equations to inequalities.

All should be inequalities.

And by the way of confirmation: Yes, (3+k)2 - 4(2)(1/2) is the discriminant for the quadratic equation that results from your original equation.
 
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  • #60
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Thank you for everyones input.
I have managed to solve this.
Can this thread be closed?
 
  • #61
jbriggs444
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Normal practice around here is to leave threads open indefinitely and simply stop posting to them.
 
  • #62
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Normal practice around here is to leave threads open indefinitely and simply stop posting to them.
Thank you @jbriggs444
 

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