How do you find the zeros, real and imaginary, of #y=x^2-3x+298# using the quadratic formula?

Question

Hazel Anglin · Accepted Answer

#x=3/2+ (13sqrt(7))/2 i#

#x=3/2- (13sqrt(7))/2 i# Standard form:#" "y=ax^2+bx+c# Where:#" " x=(-b+-sqrt(b^2-4ac))/(2a)# In your case: #a=1# #b=-3# #c=298# #=> x=(-(-3)+-sqrt((-3)^2-4(1)(298)))/(2(1))# #=> x=(3+-sqrt(-1183))/2# '~~~~~~~~~~~~~~~~~~~~~ #169xx7=1183# #169=13^2# '~~~~~~~~~~~~~~~~~~~ #=> x=(3+-sqrt(-1xx7xx13^2))/2# #=> x=(3+-13sqrt(-1)sqrt7)/2# #x=3/2+ (13sqrt(7))/2 i# #x=3/2- (13sqrt(7))/2 i#

Jameson Allen · Answer

undefined To find the zeros of the quadratic equation $y = x^2 - 3x + 298$, we use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a$, $b$, and $c$ are the coefficients of the quadratic equation $ax^2 + bx + c = 0$.

For the given equation, $a = 1$, $b = -3$, and $c = 298$. Plugging these values into the formula, we get:

$x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot 298}}{2 \cdot 1}$

$x = \frac{3 \pm \sqrt{9 - 1192}}{2}$

$x = \frac{3 \pm \sqrt{-1183}}{2}$

Since the square root of a negative number is imaginary, the zeros of the equation are:

$x = \frac{3 \pm i\sqrt{1183}}{2}$, where $i$ is the imaginary unit.