How do you find the zeros, real and imaginary, of $y = - 5 x^{2} - x + 5$ $y=-5x^2-x+5$ using the quadratic formula?

Question

1383 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-11-23T20:28:11

This equation has 2 real zeros:

$x_{1} = \frac{- 1 + \sqrt{101}}{10}$ $x_1=(-1+sqrt(101))/10$

$x_{2} = \frac{- 1 - \sqrt{101}}{10}$ $x_2=(-1-sqrt(101))/10$

To find zeros of quadratic equation you have to calculate $Delta$ first:

$Delta=b^2-4ac=(-1)^2-4*(-5)*5=1+100=101$

$Delta >0$ so the formula has 2 real zeros.

$x_1=(-b-sqrt(Delta))/(2a)=(1-sqrt(101))/(-10)=(-1+sqrt(101))/10$

$x_2=(-b+sqrt(Delta))/(2a)=(1+sqrt(101))/(-10)=(-1-sqrt(101))/10$

If $Delta<0$ then there would be 2 imaginary zeros (2 complex conjugate numbers)

If $Delta=0$ there would be a single real zero

How do you find the zeros, real and imaginary, of y=-5x^2-x+5y=−5x2−x+5y=-5x^2-x+5 using the quadratic formula?