Answer:
Option D Two irrational solutions
Step-by-step explanation:
Any quadratic equation of the form would be
\(ax^2+bx+c=0\)
We can solve this by completion of squares.
Multiply by 4a
\(4a^2x^2+4abx+4ac=0\\(2ax+b)^2-b^2+4ac=0\\(2ax+b)^2=b^2-4ac\\2ax+b=\sqrt{b^2-4ac} \\x=\frac{-b+or -\sqrt{b^2-4ac} }{2a}\)
Thus we find the solution as above
The square root if 0 we have two equal solutions
If perfect square we have two rational solutions
But here given that the discriminant b^2-4ac is positive but not perfect square
Hence the square root would be irrational thus the solution also would be irrational
Hence answer is
Option D Two irrational solutions
