Please NEED help on polynomial find possible values of k

[moeyahmed]

Please NEED help on quadratic find possible values of k

Hello
I need help I have been stuck on this question for quite a while I'm currently studying for a exam (yr 11)

The question I'm stuck with states that :

If 2kx^2-4kx+4k=4x-7 has distinct roots find the possible values of k


Any help on this question is appreciated


thank you

 

[stapel]

If 2kx^2-4kx+4k=4x-7 has distinct roots find the possible values of k

This is a quadratic:

. . . . .\(\displaystyle 2kx^2\, -\, 4kx\, +\, 4k\, =\, 4x\, -\, 7\)

. . . . .\(\displaystyle 2kx^2\, -\, 4kx\, -\, 4x\, +\, 4k\, +\, 7\, =\, 0\)

. . . . .\(\displaystyle 2k\, (x^2)\, +\, (-4k\, -\, 4)\, x\, +\, (4k\, +\, 7)\, =\, 0\)

Apply the Quadratic Formula, with:

. . . . .\(\displaystyle a\, =\, 2k,\, b\, =\, -4k\, -\, 4,\, c\, =\, 4k\, +\, 7\)

Note that, in order to have distinct (non-equal) roots, the "plus-minus" part has to be something other than zero (so that the "plus-minus" adds to and subtracts from the rest of the numerator, yielding different values). So, to find the values of "k" which will fulfill the requirements, first find the value(s) of "k" which will cause the discriminant (the insides of the square root) to be zero. Then the solution will be all other values of "k".

By the way, the above assumes that you're allowed to have complex-valued solutions. If you have to stick with real-number solutions, then you'll also need to be sure that the discriminant isn't negative, either.

If you get stuck, please reply showing your progress in following the above instructions. Thank you!

 

[moeyahmed]

Apply the Quadratic Formula, with:

. . . . .\(\displaystyle a\, =\, 2k,\, b\, =\, -4k\, -\, 4,\, c\, =\, 4k\, +\, 7\)

Note that, in order to have distinct (non-equal) roots, the "plus-minus" part has to be something other than zero (so that the "plus-minus" adds to and subtracts from the rest of the numerator, yielding different values). So, to find the values of "k" which will fulfill the requirements, first find the value(s) of "k" which will cause the discriminant (the insides of the square root) to be zero. Then the solution will be all other values of "k".

By the way, the above assumes that you're allowed to have complex-valued solutions. If you have to stick with real-number solutions, then you'll also need to be sure that the discriminant isn't negative, either.

If you get stuck, please reply showing your progress in following the above instructions. Thank you!

thank you so much staple for the quick answer
I put it into the formula but how am I supposed to solve it with k in the formula??

4k+-√16k-32k^2-28
_________________
4k


and sorry if this is a bother and again thank you so much for the help

 

[moeyahmed]

Apply the Quadratic Formula, with:

. . . . .\(\displaystyle a\, =\, 2k,\, b\, =\, -4k\, -\, 4,\, c\, =\, 4k\, +\, 7\)

Note that, in order to have distinct (non-equal) roots, the "plus-minus" part has to be something other than zero (so that the "plus-minus" adds to and subtracts from the rest of the numerator, yielding different values). So, to find the values of "k" which will fulfill the requirements, first find the value(s) of "k" which will cause the discriminant (the insides of the square root) to be zero. Then the solution will be all other values of "k".

By the way, the above assumes that you're allowed to have complex-valued solutions. If you have to stick with real-number solutions, then you'll also need to be sure that the discriminant isn't negative, either.

If you get stuck, please reply showing your progress in following the above instructions. Thank you!

thank you so much staple the help is really appreciated


I did as u told and put in to the formula but I don't know how to solve it since it has k in it I'm sorta lost


what I've done so far is

-(-4k)√-4k^2-4(2k)(4k+7)
___________________
2(2k)



so in result I ended up with this


4k√16k-32k^2-28
___________________
4k



and again thanks for the help

 

[moeyahmed]

thank you staple but still lost

thank you so much staple I really appreciate the help but I have run into a another problem see
I subbed everything into the formula but I don't know how to solve it since there k in the formula

-(-4k)√-4k^2-4(2k)(4k+7) / 2(2k)

so I ended up with this

4k√16k-32k^2-28 / 4k


thanks so much staple I really appreciate the help

 

[Deleted member 4993]

moeyahmed,

Do you know what is "discriminant" of a quadratic equation? (if not consult your textbook/Google)

What property of discriminant ensures that you have "real" roots?

What property of discriminant ensures that you have "distinct" roots?

 

[lookagain]

thank you so much staple for the quick answer I put it into the formula but how am
I supposed to solve it with k in the formula??4k+-√16k-32k^2-28_________________ 4k

You substituted wrong expressions, and it's an equation.


x = -(-4k - 4) +/- √[(-4k - 4)^2 - 4(2k)(4k + 7)]
\(\displaystyle \ \ \ \ \ \ \ \)---------------------------------------------------
\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \)2(2k)

x = 4k + 4 +/- √{[(-4)(k + 1)]^2 - 4(2k)(4k + 7)}
\(\displaystyle \ \ \ \ \ \ \ \)-----------------------------------------------------
\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \)4k

x = 4k + 4 +/- √{16(k + 1)^2 - 4(2k)(4k + 7)}
\(\displaystyle \ \ \ \ \ \ \ \)--------------------------------------------------
\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \)4k

x = 4k + 4 +/- √{4[4(k + 1)^2 - (2k)(4k + 7)]}
\(\displaystyle \ \ \ \ \ \ \ \)--------------------------------------------------
\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \)4k

x = 4k + 4 +/- 2√{4(k + 1)^2 - (2k)(4k + 7)}
\(\displaystyle \ \ \ \ \ \ \ \)-------------------------------------------------
\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \)4k

x = 2{(2k + 2 +/- √[4(k + 1)^2 - (2k)(4k + 7)]}
\(\displaystyle \ \ \ \ \ \ \ \)---------------------------------------------------
\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \)4k

x = 2k + 2 +/- √[4(k + 1)^2 - (2k)(4k + 7)]
\(\displaystyle \ \ \ \ \ \ \ \)-----------------------------------------------
\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \)2k

x = 2k + 2 +/- √[4(k^2 + 2k + 1) - (2k)(4k + 7)]
\(\displaystyle \ \ \ \ \ \ \ \)----------------------------------------------------
\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \)2k

x = 2k + 2 +/- √[4k^2 + 8k + 4 - 8k^2 - 14k]
\(\displaystyle \ \ \ \ \ \ \ \)-------------------------------------------------
\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \)2k

x = 2k + 2 +/- √[-4k^2 - 6k + 4]
\(\displaystyle \ \ \ \ \ \ \ \)-----------------------------------
\(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \)2k



But, you don't need the Quadratic Formula. But I wanted you to see a correction
on it.

You just need the discriminant, b^2 - 4ac, as was discussed prior in this thread.

 

[stapel]

thank you so much staple for the quick answer
I put it into the formula but how am I supposed to solve it with k in the formula??

4k+-√16k-32k^2-28
_________________
4k

Try following the step-by-step instructions I'd provided, which started by stating that you don't need the entire Quadratic Formula, but just the part inside the square root. What must be true of that value? What inequality or equation can you then create? And so forth.