please help me.?

Question

solve by completing the square

t^-8t+9=0

Answer 1

solving for t in
t² - 8t + 9 = 0
by completing the square.
t² - 8t = -9 by adding -9 to both sides
t² -8t + 16 = 16-9 = 7 by adding 16 to both sides.
(t-4)² = 7 by factoring the left side of the equation.
t-4 = ±√7 by taking the square root of both sides of the equation.
t = 4±√7 by adding 4 to both sides of the equation.

Or, you can use the quadratic formula, x={-b±√[b²- 4ac]}/(2a) but you'll get the same answer. I checked them... both of them... and they're correct. However, the quadratic formula is already set up. You just plug in the values. And it's derived the way we did this problem.

Answer 2

I suppose you mean t^2 - 8t + 9 = 0. You wrote the question as t^-8t + 9 = 0. That suggests t to the power -8t, plus 9 = 0

What they want you to do is to rewrite the equation in this form:
A perferct square polynomial + a constant = 0

t^2 - 8t + 9 = 0
(t)^2 - 2(t)(4) + (4)^2 - (4)^2 + 9 = 0
(t - 4)^2 - 16 + 9 = 0
(t - 4)^2 - 7 = 0
(t - 4)^2 = 7
(t - 4) = +/- (sqrt 7)
t = 4 +/- sqrt 7

Answer 3

Consider the equation :
t^2 - 8t + 9 = 0

Therefore :

t = 4 +- sqrt7

Answer 4

I am curious. What is the symbol after the first t? All I know is everything before the + needs to equal -9. So solve for t to find a solution equaling -9. Tell me what that symbol is. It's too small.

Answer 5

t^2-8t+9=0
t^2-8t=-9
{t^2-8t+[(1/2)(-8)]^2}=-9+[(1/...
{(t-4)^2}=-9+16
(t-4)^2=7
t-4=+/-[7^(1/2)]
t=+/-[7^(1/2)]+4
t=6.646 or t=1.354(up to 3 decimal places)

Answer 6

(t^2 - 2*4*t + 4^2) - 7 =0, (t-4)^2 = 7, t-4 = +-sqrt(7),

so t = 4+-sqrt(7)