How do you find the roots, real and imaginary, of #y= 5x^2 - 2x-45 # using the quadratic formula?

Question

How do you find the roots, real and imaginary, of #y= 5x^2 - 2x-45  # using the quadratic formula?

Nathan Axel · Accepted Answer

Since #b^2-4ac# is 904 which is a positive number, the given quadratic equation has real roots.

#x=(2+-sqrt904)/10# When the quadratic equation is in the form - #ax^2+bx+c=0# Then its roots are given by #x=(-b+-sqrt(b^2-(4ac)))/(2a)# In this #b^2-4ac# determines whether a given equation has real roots or imaginary roots. If #b^2-4ac# is positive the roots are positive. If #b^2-4ac# is negative the roots are negative. In our case - #(-2)^2-(4xx5xx-45)# #4 - (-900)# #4+900=904# Since #b^2-4ac# is 904 which is a positive number, the given quadratic equation has real roots. So we should be able to plug into the quadratic formula without having imaginary numbers: #x=(2+-sqrt(904))/(2(5))# #x=(2+-sqrt904)/10#

Julia Adams · Answer

undefined To find the roots of the quadratic equation $ y = 5x^2 - 2x - 45 $, you can use the quadratic formula:

$$ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} $$

where $ a = 5 $, $ b = -2 $, and $ c = -45 $.

Substitute these values into the formula:

$$ x = \frac{{-(-2) \pm \sqrt{{(-2)^2 - 4 \cdot 5 \cdot (-45)}}}}{{2 \cdot 5}} $$

$$ x = \frac{{2 \pm \sqrt{{4 + 900}}}}{{10}} $$

$$ x = \frac{{2 \pm \sqrt{{904}}}}{{10}} $$

$$ x = \frac{{2 \pm 2\sqrt{{226}}}}{{10}} $$

$$ x = \frac{1}{5} \pm \frac{{\sqrt{{226}}}}{5} $$

Therefore, the roots of the equation are:

$$ x = \frac{1}{5} + \frac{{\sqrt{{226}}}}{5} $$ and $$ x = \frac{1}{5} - \frac{{\sqrt{{226}}}}{5} $$