Question

Can somebody please help me..

Answer 1

Question please!!!!!!!

Answer 2

How many real numbers does this equation have? y+=2x^2-20x+50

Answer 3

\[\huge y+=2x^2-20x+50\]
let us compare it with
\[ \huge y+=ax^2+bx+c\] we find;

a= 2, b= -20, c= 50
Let us use quadratic formula to solve it;
\[\huge y = \frac {-b \pm \sqrt{b^2-4ac}}{2a}\]
Substituting the above values in the formula we find:
\[\huge y = \frac {-(-20) \pm \sqrt{(-20)^2-4 \times 2 \times 50}}{2 \times 2}\]
\[\huge y = \frac {20 \pm \sqrt{400-400}}{4}\]
\[\huge y = \frac {20 \pm \sqrt{0}}{4}\]
\[\huge y = \frac {20 \pm 0}{4}\]
\[\huge y = \frac {20 }{4} =5 \]
Since we hav got only one solution. hence the given eq has only one real solution i.e. real number.

@selenaannerios

Answer 4

Thats what i thought, but i wasnt sure. thank you!