a) The quadratic equation is:
[tex]25x^2 + 10 x +1 =0[/tex]
which is of the form
[tex]ax^2 + bx+c=0[/tex]
Let's analyze the radicand:
[tex]b^2-4ac=(10)^2 - 4\cdot 25 \cdot 1=100-100 =0[/tex]
The radicand is zero: this means that the equation has 2 coincident solutions.
We can find them by using the formula:
[tex]x= \frac{-b \pm \sqrt{b^2-4ac} }{2a}= \frac{-10 \pm 0}{2 \cdot 25}=-0.2 [/tex]
So, the solution is x=-0.2 with multiplicity 2.
b) The equation is
[tex]4x^2-4x+1=0[/tex]
Note that the radicand is zero again, as before:
[tex]b^2-4ac=4^2 - 4\cdot 4 \cdot 1=16-16=0[/tex]
So we have two coincident solutions as before. We can find them using the same formula:
[tex]x= \frac{-b \pm \sqrt{b^2-4ac} }{2a}= \frac{-(-4) \pm 0 }{2 \cdot 4}= \frac{4}{8}=0.5 [/tex]
So, the solution is x=0.5 with multiplicity 2.
