How do you find the roots, real and imaginary, of # h=--t^2+16t -4 # using the quadratic formula?

Question

How do you find the roots, real and imaginary, of # h=--t^2+16t -4  # using the quadratic formula?

Eva Abramson · Accepted Answer

#color(blue)(h=-8+2sqrt17 or h=-8-2sqrt17# #h=- -t^2+16t-4# #h=-(-t^2)+16t-4# #h=t^2+16t-4# #x=(-b+-sqrt(b^2-4ac))/(2a)# #color(blue)(a^2+bx+c# #color(blue)(a=1,b=16,c=-4# #h=(-(16)+-sqrt((16)^2-4(1)(-4)))/(2(1))# #(-16+-sqrt(256+16))/2# #(-16+-sqrt(272))/2# #(-16+-sqrt(2*2*2*2*17))/2# #color(blue)(sqrt2 xx sqrt2=2# #(-16+-4sqrt17)/2# #(cancel4^color(blue)2(-4+sqrt17))/cancelcolor(blue)2^color(blue)1 or (cancel4^color(blue)2(-4-sqrt17))/cancel2^color(blue)1 # #2(-4+sqrt17) or 2(-4-sqrt17)) # #color(blue)(-8+2sqrt17 or -8-2sqrt17#

Nathan Axel · Answer

undefined To find the roots of the quadratic equation $h = -t^2 + 16t - 4$ using the quadratic formula, $t = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$, where $a = -1$, $b = 16$, and $c = -4$:

1. Substitute the values of $a$, $b$, and $c$ into the quadratic formula: $t = \frac{{-16 \pm \sqrt{{16^2 - 4(-1)(-4)}}}}{{2(-1)}}$.

2. Simplify inside the square root: $t = \frac{{-16 \pm \sqrt{{256 - 16}}}}{{-2}}$.

3. Further simplify inside the square root: $t = \frac{{-16 \pm \sqrt{{240}}}}{{-2}}$.

4. Evaluate the square root: $\sqrt{{240}} \approx 15.49$.

5. Substitute the value of the square root into the equation: $t = \frac{{-16 \pm 15.49}}{{-2}}$.

6. Solve for both roots: $t_1 = \frac{{-16 + 15.49}}{{-2}}$ and $t_2 = \frac{{-16 - 15.49}}{{-2}}$.

7. Calculate the values: $t_1 \approx 0.255$ and $t_2 \approx 15.745$.

Therefore, the roots of the equation $h = -t^2 + 16t - 4$ are approximately $t \approx 0.255$ and $t \approx 15.745$.

Hazel Anglin · Answer

undefined To find the roots of the quadratic equation h = -t^2 + 16t - 4 using the quadratic formula, you first identify the coefficients a, b, and c. Then, you substitute these values into the quadratic formula:

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

where:
- $ a = -1 $
- $ b = 16 $
- $ c = -4 $

Substitute these values into the quadratic formula and solve for t. This will give you the roots, which may be real or imaginary depending on the discriminant (the value under the square root).