How do you find the zeros, real and imaginary, of #y= -7x^2+4x+2 # using the quadratic formula?

Question

Nathan Axel · Accepted Answer

#x = 2/7+3/7sqrt(2)" "# or #" "x = 2/7-3/7sqrt(2)#

#y = -7x^2+4x+2#

is in the form:

#y = ax^2+bx+c#

with #a=-7#, #b=4# and #c=2#

This has zeros given by the quadratic formula:

#x = (-b+-sqrt(b^2-4ac))/(2a)#

#color(white)(x) = (-color(blue)(4)+-sqrt(color(blue)(4)^2-4(color(blue)(-7))(color(blue)(2))))/(2(color(blue)(-7)))#

#color(white)(x) = (-4+-sqrt(16+56))/(-14)#

#color(white)(x) = (-4+-sqrt(72))/(-14)#

#color(white)(x) = (-4+-sqrt(6^2*2))/(-14)#

#color(white)(x) = (-4+-6sqrt(2))/(-14)#

#color(white)(x) = (-2+-3sqrt(2))/(-7)#

#color(white)(x) = 2/7+-3/7sqrt(2)#

That is:

#x = 2/7+3/7sqrt(2)" "# or #" "x = 2/7-3/7sqrt(2)#

Ethan Adkins · Answer

undefined To find the zeros of the quadratic equation $y = -7x^2 + 4x + 2$, you can use the quadratic formula, which is:

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

For the given equation, $a = -7$, $b = 4$, and $c = 2$.

Substitute these values into the quadratic formula:

$$x = \frac{{-4 \pm \sqrt{{4^2 - 4(-7)(2)}}}}{{2(-7)}}$$

$$x = \frac{{-4 \pm \sqrt{{16 + 56}}}}{{-14}}$$

$$x = \frac{{-4 \pm \sqrt{{72}}}}{{-14}}$$

$$x = \frac{{-4 \pm 6\sqrt{2}i}}{{-14}}$$

Therefore, the zeros of the equation are:

$$x_1 = \frac{{-4 + 6\sqrt{2}i}}{{-14}}$$

$$x_2 = \frac{{-4 - 6\sqrt{2}i}}{{-14}}$$