x^(2)-18x+81=0?

Question

i cant figure this out someone please help me. i am studying for a algebra 2 final and its on monday! please help!

Answer 1

Solve for "x"...

x^2 - 18x + 81 = 0

First: multiply the 1st & 3rd term to get 81. find two numbers that give you 81 when multiplied & -18 (2nd term) when added/subtracted. the numbers are (-9 & -9). rewrite the expression with the new middle terms.

x^2 - 9x - 9x + 81 = 0

Sec: group "like" terms; factor both sets of parenthesis.

(x^2 - 9x) - (9x + 81) = 0
x(x-9) - 9(x-1) = 0
(x-9)(x-9) = 0
(x-9)^2 = 0

Third: eliminate the exponent - find the square root of both sides.

V`(x-9)^2 = V`0
x - 9 = 0 (add 9 to both sides - when you move a term to the opposite side, always use the opposite sign).

x - 9+9 = 0+9
x = 9

Answer 2

x^2 - 18x +81 = 0
(x-9)(x-9) = 0
Therefore, x = 9

Answer 3

As written: x² = -80 one ?x² =?(-80 one) = ?80 one* ?(-a million) bear in mind ?(-a million) = i x= ±9i yet once you meant: x² + 18x = -80 one x² + 18x + 80 one =0 (x + 9)² = 0 x - 9 = 0 x = 9 it particularly is a double root(answer)

Answer 4

x² - 18x + 81 = 0
(x - 9).(x - 9) = 0
x = 9 (twice)

Answer 5

x^(2)-18x+81=0
(x-9) * (x-9) = 0

x = 9

Answer 6

x^(2) -18x+81=0
(x -9)*(x-9) = 0
x = 9 (double root)

Answer 7

two numbers that multiply to 81 and add to -18

the two numbers are -9 and -9

(x - 9) (x - 9) = 0 or (x-9)^2 = 0
x = 9

Answer 8

x^2 - 18x + 81 = (x - 9)^2, so now it's boringly simple to find out the root..
Regards
Tonio

Answer 9

factor into (x-9)(x-9)=0, a number times 0 equals 0 so you know that x-9=0, and x equals 9.

Answer 10

its a quadratic equation so use the quatratic equation, u should of learn or easily find on line