How do you find the zeros of # y = 2x^2 + 4x -1 # using the quadratic formula?

Question

How do you find the zeros of # y =  2x^2 + 4x -1 # using the quadratic formula?

Hazel Anglin · Accepted Answer

#-1+-sqrt(6)/2#

This quadratic function is written in "standard form", or #ax^2+bx+c#. The quadratic formula uses the numbers a, b, and c like this: #(-b+-sqrt(b^2-4ac))/(2a)# comparing the original function to the general, standard form function above, we see that #a=2, b=4, &c=-1#.

Plug these in the quadratic formula: #(-4+-sqrt(4^2-4(2)(-1)))/(2*2)=(-4+-sqrt(16-(-8)))/4# #(-4+-sqrt(24))/4# To simplify that #sqrt(24)#, pull out the largest perfect square that goes in it. Since 4 divides evenly into 24, and neither 9 nor 16 do, break it down in terms of 4: #sqrt(24)=sqrt(4*6)# You can take out #sqrt(4)=2#:

#sqrt(4*6)=2sqrt(6)# From here, divide through by 4 to get #-1+-sqrt(6)/2#

...and remember, quadratic functions always have two solutions!

Isaiah Anthony · Answer

undefined To find the zeros of $y = 2x^2 + 4x - 1$ using the quadratic formula, first identify the coefficients:

$a = 2$
$b = 4$
$c = -1$

Then, apply the quadratic formula:

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

Substitute the coefficients into the formula:

$$x = \frac{{-4 \pm \sqrt{{4^2 - 4 \cdot 2 \cdot (-1)}}}}{{2 \cdot 2}}$$

Simplify inside the square root:

$$x = \frac{{-4 \pm \sqrt{{16 + 8}}}}{{4}}$$

$$x = \frac{{-4 \pm \sqrt{{24}}}}{{4}}$$

$$x = \frac{{-4 \pm 2\sqrt{{6}}}}{{4}}$$

$$x = \frac{{-1 \pm \sqrt{{6}}}}{{2}}$$

Therefore, the zeros of the quadratic equation are:

$$x = \frac{{-1 + \sqrt{{6}}}}{{2}}$$ and $$x = \frac{{-1 - \sqrt{{6}}}}{{2}}$$