solve using the quadratic formula: x^2 - 3x - 3 = -5?

Answer 1

x² - 3x - 3 = - 5

x² - 3x - 3 = 5 = - 5 + 5

x² - 3x + 2

- - - - - - -

a = 1

b = - 3,

c = 2

- - - - - - - - - - --

Quadratic formula

x = -b ± √b² - 4ac/ 2a

x = - (- 3) ± √(- 3)² - 4(1)(2)

x = 3 ± √9 - 8 /2

x = 3 / 2 ± √1 / 2

x = 1 1/2 ± 1/2

x = 1.5 ± 0.5

- - - - - - - - -

Solving for +

x = 1.5 + 0.5

x = 2

- - - - - - - - - -

Solving for -

x = 1.5 - 0.5

x = 1

- - - - - - - - -s-

Answer 2

x^2 - 3x -3 = -5
implies x^2 - 3x +2 = 0
Method-1: Factorisation
x^2 - 2x - x +2 = 0
x(x-2)-1(x-2) = 0
or (x-2)(x-1)=0
now, either x-2=0 or x-1=0
or x=2 or x=1

Method-2: Quadratic formula
x = (-b+ sqrt (b*b-4ac) )/2a
x = (-b - sqrt(b*b- 4ac))/2a
where a,b anc c are constants in the equation ax^2+bx+c=0
in the problem x*x - 3x + 2 =0,
b= -3
a= 1
c= 2
using the formulae for x we have:
x = (3+sqrt(9-8))/2 = 2
x = (3- sqrt(9-8))/2 = 1

Answer 3

Add 5 from both sides:
x^2-3x+2=0

(x-1)(x-2)=0
x=1 or 2

Check:
(1)^2-3(1)+2=0
1-3+2=0
-2+2=0
0=0

(2)^2-3(2)+2=0
4-6+2=0
-2+2=0
0=0

Answer 4

Add 5 to each side, x*2-3x+2=0.
This becomes (x-2)(x-1) = 0
x-2=0, x=2
x-1=0, x=1

Answer 5

x^2 - 3x + 2 = 0

(x - 1)(x - 2) = 0

x = 1 or 2

Answer 6

hi , i think of you the formula and you shall get the respond.enable's attempt to derive an formula for a well-known quadratic words . enable us to take a quadratic equation ax^2+bx+c = 0 now we ought to derive the value of "x". If p*q = 0 then we get the two p = 0 OR q = 0. So if i'm getting in terms of p,q we get our answer.So enable us to purpose. ------> Multiply via 4a on the two aspects we get (2ax)^2+2(2ax)(b)+4ac = 0. ------->Now enable us to purpose to get interior the formula of (a+b)^2.Addind and subtracting b^2 on the left hand area we get (2ax)^2+2(2ax)(b)+b^2--b^2+4ac = 0 --------> Now I write the 1st 3 words as ( 2ax+b)^2 and take "--" person-friendly interior the 2nd so i'm getting a formula of p^2--q^2 which may well be written interior certainly one of those (p+q)(p--q) ---------> ( 2ax+b)^2--{sqrt(b^2--4ac)}^2 = 0 ---------> {2ax+b+sqrt(b^2--4ac)} {2ax+b--sqrt(b^2--4ac)} = 0 ----------> x = [--b+(or)--{sqrt(b^2--4ac)}]/2a word:This formula is ideal whan quadratic equation is in time-honored kind. Now enable us to resolve your subject taking a = a million b = --3 c = --8 (via reworking 5 on the different area) save on with the above formula x = [--(--3)+(or)--{sqrt((--3)^2--4(a million)(--8)}... via fixing you get x = [3+(or)--sqrt41]/2 answer: x = [3+(or)--sqrt41]/2

Answer 7

x^2-3x-3+5=0
x^2-3x+2=0
d= b^-4ac
d=9-4*2*2
d=9-16
d=-7
discriminant is less than 0 so no quad. equation will form

Answer 8

x^2 - 3x - 3 = -5
x^2 - 3x +2 = 0

A=1
B=-3
C=2
D=B^2-4*A*C = (-3)^2-4*1*2 = 9-8 = 1
sqrt(D) = +/-1

x=[-B+/-sqrt(D)]/2*A
x = (3 +/-1)/2*1
x =2 or 1
Roots are thus x=2 and x=1

Answer 9

x^2 -3x+2=0
x^2 -x-2x+2=0
x(x-1)-2(x-1)=0
x=1orx=2

Answer 10

Good Job Class! You all passed the test! Now do my homework for me!