How do you use the quadratic formula to solve #16t^2-4t+3=0#?

Question

Lincoln Anderson · Accepted Answer

#t = 1/8+-1/8sqrt(11)i#

Given:

#16t^2-4t+3 = 0#

This is a quadratic equation in standard form:

#at^2+bt+c = 0#

with #a=16#, #b=-4# and #c=3#

Its roots can be found using the quadratic formula:

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

#color(white)(t) = (-(color(blue)(-4))+-sqrt((color(blue)(-4))^2-4(color(blue)(16))(color(blue)(3))))/(2(color(blue)(16)))#

#color(white)(t) = (4+-sqrt(16-192))/32#

#color(white)(t) = (4+-sqrt(-176))/32#

#color(white)(t) = (4+-sqrt(176)i)/32#

#color(white)(t) = (4+-sqrt(16*11)i)/32#

#color(white)(t) = (4+-4sqrt(11)i)/32#

#color(white)(t) = 1/8+-1/8sqrt(11)i#

Liam Brennan · Answer

undefined To use the quadratic formula to solve $16t^2 - 4t + 3 = 0$, follow these steps:

1. Identify the coefficients $a$, $b$, and $c$ in the quadratic equation $ax^2 + bx + c = 0$.
2. Substitute the values of $a$, $b$, and $c$ into the quadratic formula: $x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$.
3. Calculate the discriminant, $b^2 - 4ac$, to determine the nature of the roots.
4. If the discriminant is positive, there are two distinct real roots. If it's zero, there is one real root (a repeated root). If it's negative, there are two complex roots.
5. Substitute the values of $a$, $b$, and $c$ into the quadratic formula and simplify to find the values of $t$.

For the equation $16t^2 - 4t + 3 = 0$:
- $a = 16$
- $b = -4$
- $c = 3$

Substitute these values into the quadratic formula:
$$ t = \frac{{-(-4) \pm \sqrt{{(-4)^2 - 4(16)(3)}}}}{{2(16)}} $$

Calculate the discriminant:
$$ b^2 - 4ac = (-4)^2 - 4(16)(3) = 16 - 192 = -176 $$

Since the discriminant ($-176$) is negative, there are two complex roots.

Substitute the values of $a$, $b$, and $c$ into the quadratic formula and simplify:
$$ t = \frac{{4 \pm \sqrt{{-176}}}}{{32}} $$
$$ t = \frac{{4 \pm \sqrt{{176i}}}}{{32}} $$
$$ t = \frac{{4 \pm 4\sqrt{{11}}i}}{{32}} $$
$$ t = \frac{{1 \pm \sqrt{{11}}i}}{{8}} $$

Therefore, the solutions to the equation $16t^2 - 4t + 3 = 0$ are $t = \frac{{1 + \sqrt{{11}}i}}{{8}}$ and $t = \frac{{1 - \sqrt{{11}}i}}{{8}}$, where $i$ is the imaginary unit.