How do you find the zeros, real and imaginary, of #y=-2x^2-9x+5# using the quadratic formula?

Question

Abigail Allen · Accepted Answer

Zeros are real and #x=0.5, x= -5# # y= -2 x^2-9 x +5# Comparing with standard quadratic equation #ax^2+bx+c=0# # a=-2 ,b=-9 ,c=5#. Discriminant # D= b^2-4 a c# or #D=81+40 =121# , discriminant is positive, we get two real solutions. Quadratic formula: #x= (-b+-sqrtD)/(2a) #or #x= (9+-sqrt 121)/(-4) = (9 +- 11)/-4 :. x= 20/-4= -5 # or #x=(-2)/-4 = 0.5 :. # Zeros are real and #x=0.5, x= -5# [Ans]

Jameson Allen · Answer

undefined To find the zeros of the quadratic function $ y = -2x^2 - 9x + 5 $, you can use the quadratic formula, which is given by:

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

where $ a = -2 $, $ b = -9 $, and $ c = 5 $.

Plugging these values into the quadratic formula:

$$ x = \frac{{-(-9) \pm \sqrt{{(-9)^2 - 4(-2)(5)}}}}{{2(-2)}} $$

$$ x = \frac{{9 \pm \sqrt{{81 + 40}}}}{{-4}} $$

$$ x = \frac{{9 \pm \sqrt{{121}}}}{{-4}} $$

$$ x = \frac{{9 \pm 11}}{{-4}} $$

So, the zeros are:

$$ x = \frac{{9 + 11}}{{-4}} $$ and $$ x = \frac{{9 - 11}}{{-4}} $$

$$ x = \frac{{20}}{{-4}} $$ and $$ x = \frac{{-2}}{{-4}} $$

$$ x = -5 $$ and $$ x = \frac{1}{2} $$

Therefore, the real zeros are $ x = -5 $ and $ x = \frac{1}{2} $, and there are no imaginary zeros.