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

  • Thread starter Jaco Viljoen
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In summary, the given quadratic equation has no real roots if the discriminant, k^2 + 6k + 5, is less than zero. To find the values of k that make this true, we need to solve the inequality k^2 + 6k + 5 < 0. Using the quadratic formula, we get the solutions k = -1 and k = -5. Therefore, the values of k that make the discriminant negative are -1 and -5, which means that the original quadratic equation has no real roots when k = -1 or k = -5.
  • #1
Jaco Viljoen
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9

Homework Statement


2x^2-3x+kx=-1/2

1. k<1 or k>1
2. 1<=k<=5
3. k<=1 or k>=5
1<k<5

Homework Equations


b^2-4ac
a=2 b=3 c=k

The Attempt at a Solution


(3)^2-4(2)(k)
=9-8k<0
=9/8<k
=1&1/8<k

I get the answer above but don't know how it relates?
Any insight would be appreciated.

Thank you,
Jaco
 
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  • #2
Find the discriminant of your quadratic equation. The equation has no real roots if the discriminant is less than zero.
 
  • #3
Jaco Viljoen said:

Homework Statement


2x^2-3x+kx=-1/2

Homework Equations


b^2-4ac
a=2 b=3 c=k
Look at the original equation again.

What is the coefficient on x? What is the constant term?
 
  • #4
Are you sure about your value for b?
 
  • #5
Jbriggs,
I have been looking at other threads and found a similar example and there:
(-3)^2-4(2)(k+1/2)
=-9-8k+4
=-5-8k
=> k=5/8
Thank you for pointing that out Marcus
 
  • #6
Is this correct?
 
  • #7
It doesn't look correct. What is the general form of a quadratic equation? Write your equation in a way that imitates that general form and identify correctly the values of a,b,c. Then write down the general definition of the discriminant, and substitute your values of a,b,c into it. The solve the inequality (discriminant) < 0 for k.
 
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  • #8
Jaco Viljoen said:

Homework Statement


2x^2-3x+kx=-1/2

1. k<1 or k>1
2. 1<=k<=5
3. k<=1 or k>=5
1<k<5

Homework Equations


b^2-4ac
a=2 b=3 c=k

The Attempt at a Solution


(3)^2-4(2)(k)
=9-8k<0
=9/8<k
=1&1/8<k

I get the answer above but don't know how it relates?
Any insight would be appreciated.

Thank you,
Jaco
Please restate the entire problem as it was given to you.

You may have a typo in the quadratic expression as you posted it. If it's correct as it is, then you still do not have the correct b or c .
 
  • #9
Sammy,
For what values of k will the equation 2x^2 -3x + kx = -1/2 have no real roots?

Possible answers:
1. k<1 or k>1
2. 1<=k<=5
3. k<=1 or k>=5
1<k<5
 
  • #10
Everyone responding to this thread is attempting to point out that the b and c that you have harvested from that equation are wrong.
 
  • #11
ok, I have been considering this too as:
(3x+kx)^2-4(2)(-1/2)
(3x+kx)(3x+kx)+4
9x^2+3kx^2+3kx^2+k^2x^2
9x^2+6kx^2+k^2x^2
 
  • #12
Re-read post #7 above. What is the standard form for a quadratic equation? Can you restate the original equation in that form?
 
  • #13
ax^2+bx+c=0
2x^2+(3x+kx)+1/2=0
2x^2+3x+kx=-1/2
kx=-1/2-2x^2-3x
k=(1/2-2x^2-3x)/x
 
  • #14
wow, i feel more confused...
 
  • #15
Jaco Viljoen said:
ax^2+bx+c=0
2x^2+(3x+kx)+1/2=0
It's good to this point.

Take the expression in parentheses and factor out x.
 
  • #16
2x^2+(3x+kx)+1/2=0
2x^2+x(3+k)+1/2=0
2x^2+(3+k)(x+1/2)=0
2x^2+3x+1&1/2+kx+1/2k=0
kx+1/2k=-2x^2-3x-1&1/2
3/2kx=-2x+3+1&1/2
3kx=-4x+9
k=(-4x+9)/x
 
Last edited:
  • #17
Jaco Viljoen said:
2x^2+(3x+kx)+1/2=0
2x^2+x(3+k)+1/2=0
Stop at this point !
2x^2+(3+k)(x+1/2)=0
2x^2+3x+1&1/2+kx+1/2k=0
kx+1/2k=-2x^2-3x-1&1/2
1&1/2k=(-2x+3+1&1/2)/x
What is the coefficient of x ?
 
  • #18
1?
 
  • #19
Jaco Viljoen said:
1
In the expression 2x2 + 3x + 4, what is the coefficient on the "x" term?
 
  • #20
3
 
  • #21
Good.

Now, in the expression 2x2 + x(3+k) + 1/2, what is the coefficient on the x term?
 
  • #22
2x^2+1x(3+k)+1/2

1 or is it still 3?

2x^2+1x(3+k)+1/2
 
  • #23
I am missing this, I just can't seem to get it...
 
  • #24
Its not the 2 is it?
 
  • #25
What if we write it like this?

2x2 + (3+k)x + (1/2) = 0

What is the coefficient of x2 ?

What is the coefficient of x ?

What is the constant term ?
 
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  • #26
2
3+k
1/2
 
  • #27
Jaco Viljoen said:
2
3+k
1/2
Now proceed.
 
  • #28
x^2+(3+k)+1/2=0
like this?
 
  • #29
x^2+(3+k)+1/2=0
a b c
x = (-(3+k) +/-√((3+k)2 - 4x^2(1/2)))/2x^2
 
  • #30
Jaco Viljoen said:
x^2+(3+k)+1/2=0
a b c
x = (-(3+k) +/-√((3+k)2 - 4x^2(1/2)))/2x^2

In the discriminant you where you are trying to write b^2 - 4ac. What was the value you came up with for a again?
 
  • #31
(x+ )(x+ )=0
 
  • #32
Jaco Viljoen said:
(x+ )(x+ )=0
No. That is not correct and is not what SammyS agreed was correct. Refer back to your post #26.
 
  • #33
jbriggs,
I don't understand what you are asking?
 
  • #34
2
3+k
1/2
 
  • #35
Jaco Viljoen said:
2
3+k
1/2
That's a = 2, b = 3+k and c=1/2

Given that, what is the value of a?
 

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