Hello everyone,
I solved this quadratic equation using logic, but my math teacher said that it could also be solved using the Quadratic Formula. I do not how to do the latter because it seems that the discriminant would become negative (Imaginary numbers have not been taught yet to the class, so the question shouldn't involve them).
Any help would be appreciated!
Thank you.
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1. Solve the following for x in terms of y:
\(\displaystyle x^2 - 2x + 1 + y^2 = 0\)
My method :
\(\displaystyle (x - 1)^2 + y^2 = 0\)
\(\displaystyle x = 1\), because for the sum of any two positive numbers to be zero, the two positive numbers must be 0 themselves.
Teacher's method :
a = 1, b = -2, c = \(\displaystyle y^2 + 1\)
Therefore, the quadratic formula would yield:
\(\displaystyle x = \frac{2 \pm \sqrt{4 - 4(y^2 + 1)}}{2}\)
\(\displaystyle x = \frac{2 \pm \sqrt{4 - 4y^2 - 4}}{2}\)
\(\displaystyle x = 1 \pm \frac{ \sqrt{- 4y^2}}{2}\)
I solved this quadratic equation using logic, but my math teacher said that it could also be solved using the Quadratic Formula. I do not how to do the latter because it seems that the discriminant would become negative (Imaginary numbers have not been taught yet to the class, so the question shouldn't involve them).
Any help would be appreciated!
Thank you.
---
1. Solve the following for x in terms of y:
\(\displaystyle x^2 - 2x + 1 + y^2 = 0\)
My method :
\(\displaystyle (x - 1)^2 + y^2 = 0\)
\(\displaystyle x = 1\), because for the sum of any two positive numbers to be zero, the two positive numbers must be 0 themselves.
Teacher's method :
a = 1, b = -2, c = \(\displaystyle y^2 + 1\)
Therefore, the quadratic formula would yield:
\(\displaystyle x = \frac{2 \pm \sqrt{4 - 4(y^2 + 1)}}{2}\)
\(\displaystyle x = \frac{2 \pm \sqrt{4 - 4y^2 - 4}}{2}\)
\(\displaystyle x = 1 \pm \frac{ \sqrt{- 4y^2}}{2}\)
!