Solving a quadratic equation with k as a prefactor?
Hello dear people,
I've been given a quadratic equation (see image) with k as a prefactor instead of a fixed number. The abc formula and pq formula are to be applied, and the solution has also been given (second image), but despite lots of trial and error, I can't find the required solution. How do I handle these letters as prefactors? Can anyone give me a hint? Unfortunately, I couldn't solve it by simply substituting them into the formula, but I probably solved for x incorrectly.
Thank you very much and have a nice evening
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You do exactly the same steps as always, just think that k is a number, just write as a letter. To apply the pq formula, you must first get the 2 away from the x^2 (because the pq formula is to be used only in equations of the form x^2+px+q=0). So you are sharing through 2 and instead
x^2 – 0.5kx – 0.5k^2 = 0
Now you only need p and q you use in the pq formula. You can actually read this now.
p = -0.5k (as in the middle -0.5kx and that would be the px part)
q=0.5k^2
The pq formula is x= -0.5p ± root(0,25p^2 – q)
^ Yes, I have rewritten the pq formula a bit because I write on the phone and that looks funny with 1/2 and many clips
With your values inserted in the formula you will get (see sign rules like minus = plus)
x1 = 0.25k – root (0.25*(-0.5k)^2 + 0.5k^2)
The x1 = 0.25k – root (0,625k^2) or
x1 = 0.25k – 0.75k = -0.5k
Thus x2 = 0.25k + 0.75k = k
If there are any questions, please ask them under my answer. I think I’ve exchanged x1 and x2 in my answer, but that’s not bad, it’s just the two solutions.
Thanks for the super detailed answer, now I understand it 🙂 Will the sign of -0.125k^2 be ignored when the values are inserted into the formula because it is already squared and correspondingly positive? Because otherwise you could not add it so easily with 0.5k^2 and get 0.625k^2. I always had 0.375k^2 out there in my bill because I thought that was subtracted.
Yes in this case it falls away because 0.25*(-0.5k)^2 is the same as 0.25*0.25k^2 = 0.125k^2. So, as you said, the minus falls away because of cubbering
What’s your way of calculating?
For ABC formula simply use a=2, for b= -k and for c= -k2.
Thus you get under the root to 9k2
Thanks for the tip!