How do you find the zeros, real and imaginary, of #y=-4x^2-2x+3 #using the quadratic formula?

Question

Hazel Anglin · Accepted Answer

#(-1+--1sqrt(13))/4#

The quadratic formula is #-b/(2a)+-sqrt(b^2-4*a*c)/(2a)#

Let's look at our equation, #-4x^2-2x+3#, in a more general form: #ax^2-bx+c#.

So #a# is #-4# #b# is #-2# #c# is #3#.

Now let's plug our variables into the quadratic formula. I like to put parentheses around the variables I'm replacing, just to be extra careful about silly little sign errors.

#(-(-2))/(2(-4))+-sqrt((-2)^2-4*(-4)*(3))/(2(-4))#.

This becomes #2/-8+-sqrt(4-(-48))/-8#, which we can simplify to #-1/4+-(2sqrt(13))/-8#, or #-1/4+-(-1sqrt(13))/4#.

The final answer is #(-1+--1sqrt(13))/4#

Kennedy Adams · Answer

undefined The quadratic formula states that for a quadratic equation of the form $ax^2 + bx + c = 0$, the solutions are given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. For the equation $y = -4x^2 - 2x + 3$, $a = -4$, $b = -2$, and $c = 3$. Plugging these values into the formula, we get:

$x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(-4)(3)}}{2(-4)}$

$x = \frac{2 \pm \sqrt{4 + 48}}{-8}$

$x = \frac{2 \pm \sqrt{52}}{-8}$

The square root of 52 can be simplified to $2\sqrt{13}$. So, the solutions are:

$x = \frac{2 \pm 2\sqrt{13}}{-8}$

$x = \frac{1 \pm \sqrt{13}}{-4}$

Therefore, the zeros of the equation are $x = \frac{1 + \sqrt{13}}{-4}$ and $x = \frac{1 - \sqrt{13}}{-4}$, which are both real and irrational.