(x-3)^2 - (x-3) -2 = 0
Do i just find what x is???
Solve??? MATH PROBLEM!!!!~~~~~~~?
2008-12-22 3:11 am
回答 (30)
2008-12-22 3:31 am
✔ 最佳答案
let y = x - 3, thenthe given eguation can be written as:
y^2 - y - 2 = 0; or
y^2 + y - 2y - 2 = 0 (replace -y = +y - 2y)
y(y + 1) - 2(y + 1) = 0 (group factor)
(y + 1)( y - 2) = 0; or
[(x - 3) + 1][(x - 3) - 2] = 0
(x - 2)(x - 5) = 0 (replace back y = x - 3)
Therefore:
x - 2 = 0 ==> x = 2
x - 5 = 0 ==> x = 5
The solutions are 2 or 5.
2008-12-22 7:12 pm
(x - 3)² - (x - 3) - 2 = 0
x² - 3x - 3x + 9 - x + 3 = 2
x² - 7x = - 10
x² - 7/2x = - 10 + (- 7/2)²
x² - 7/2x = - 40/4 + 49/4
(x - 7/2)² = 9/4
x - 7/2 = 3/2
x = 3/2 + 7/2, x = 10/2, x = 5
x = - 3/2 + 7/2, x = 4/2, x = 2
Answer: x = 5, 2
Proof (x = 5):
= (5 - 3)² - (5 - 3) - 2
= 2² - (2) - 2
= 4 - 2 - 2
= 0
Proof (x = 2):
= (2 - 3)² - (2 - 3) - 2
= (- 1)² - (- 1) - 2
= 1 + 1 - 2
= 0
x² - 3x - 3x + 9 - x + 3 = 2
x² - 7x = - 10
x² - 7/2x = - 10 + (- 7/2)²
x² - 7/2x = - 40/4 + 49/4
(x - 7/2)² = 9/4
x - 7/2 = 3/2
x = 3/2 + 7/2, x = 10/2, x = 5
x = - 3/2 + 7/2, x = 4/2, x = 2
Answer: x = 5, 2
Proof (x = 5):
= (5 - 3)² - (5 - 3) - 2
= 2² - (2) - 2
= 4 - 2 - 2
= 0
Proof (x = 2):
= (2 - 3)² - (2 - 3) - 2
= (- 1)² - (- 1) - 2
= 1 + 1 - 2
= 0
2008-12-22 11:14 am
Yes first multiply the x-3^2 out and put the like terms together the x-3 and then subtract 2
2008-12-22 2:01 pm
(x - 3)^2 - (x - 3) - 2 = 0
Change (x - 3) into y:
(x - 3)^2 - (x - 3) - 2 = 0
y^2 - y - 2 = 0
y^2 + y - 2y - 2 = 0
(y^2 + y) - (2y + 2) =
y(y + 1) - 2(y + 1) = 0
(y + 1)(y - 2) = 0
Change y into (x - 3):
(y + 1)(y - 2) = 0
[(x - 3) + 1][(x - 3) - 2] = 0
[x - 3 + 1][x - 3 - 2] = 0
[x - 2][x - 5] = 0
x - 2 = 0
x = 2
x - 5 = 0
x = 5
â´ x = 2 , 5
Change (x - 3) into y:
(x - 3)^2 - (x - 3) - 2 = 0
y^2 - y - 2 = 0
y^2 + y - 2y - 2 = 0
(y^2 + y) - (2y + 2) =
y(y + 1) - 2(y + 1) = 0
(y + 1)(y - 2) = 0
Change y into (x - 3):
(y + 1)(y - 2) = 0
[(x - 3) + 1][(x - 3) - 2] = 0
[x - 3 + 1][x - 3 - 2] = 0
[x - 2][x - 5] = 0
x - 2 = 0
x = 2
x - 5 = 0
x = 5
â´ x = 2 , 5
2008-12-22 11:28 am
yes. when the equation is equal to 0, you must solve the value of a variable.
heres the solution
(x-3)^2-(x-3)-2=0
(x-3)(x-3)-x+3-2=0
x^2-6x+9-x+1=0
x^2-7x+10=0
(x-5)(x-2)=0
x-5=0 x-2=0
x=5 x=2 answer
I hope this helps
heres the solution
(x-3)^2-(x-3)-2=0
(x-3)(x-3)-x+3-2=0
x^2-6x+9-x+1=0
x^2-7x+10=0
(x-5)(x-2)=0
x-5=0 x-2=0
x=5 x=2 answer
I hope this helps
2008-12-22 11:27 am
First, you unfactor (x-3)^2 and get, x^2 - 6x + 9. Then, you combine "like" terms and get:
x^2 - 6x + 9 - x + 3 -2 = 0
x^2 - 7x + 10 = 0
Then just factor this.
(x-5)(x-2)
X= {2,5}
tHANX STUDENT XD ALSO THANX FOR THE 2 POINTS LOL
x^2 - 6x + 9 - x + 3 -2 = 0
x^2 - 7x + 10 = 0
Then just factor this.
(x-5)(x-2)
X= {2,5}
tHANX STUDENT XD ALSO THANX FOR THE 2 POINTS LOL
2008-12-22 11:24 am
Yes, I think in this case X = 6.372 or 0.628
(x-3)^2 - (x-3) -2 = 0
(X-3)(X-3)-X-5 = 0
X^2-3X-3X+9-X-5 = 0 <---Multiply out using FOIL
X^2-7X+4 = 0
You can use the quadratic equation to solve this
I don't know how to do square root symbols and everything on a computer, but by doing it on paper it looks like your answer will be 3.5 plus or minus the square root of 33 over 2.
(x-3)^2 - (x-3) -2 = 0
(X-3)(X-3)-X-5 = 0
X^2-3X-3X+9-X-5 = 0 <---Multiply out using FOIL
X^2-7X+4 = 0
You can use the quadratic equation to solve this
I don't know how to do square root symbols and everything on a computer, but by doing it on paper it looks like your answer will be 3.5 plus or minus the square root of 33 over 2.
2008-12-22 11:16 am
yes
x^2 - 6x +9 - x +3 -2 =0
x^2-7x+10=0
factor
(x-5)*(x-2)
x=5 and 2
x^2 - 6x +9 - x +3 -2 =0
x^2-7x+10=0
factor
(x-5)*(x-2)
x=5 and 2
2008-12-22 11:21 am
You could let y = x - 3, so that
y^2 - y - 2 = 0
(y - 1/2)^2 - 2 - 1/4 = 0
(completing the square)
(y - 1/2)^2 = 9/4
y = 1/2 (+-)3/2
so if y = x - 3, x = y + 3
so x = 7/2 (+-)3/2
x = 4/2 = 2 or x = 10/2 = 5.
y^2 - y - 2 = 0
(y - 1/2)^2 - 2 - 1/4 = 0
(completing the square)
(y - 1/2)^2 = 9/4
y = 1/2 (+-)3/2
so if y = x - 3, x = y + 3
so x = 7/2 (+-)3/2
x = 4/2 = 2 or x = 10/2 = 5.
2008-12-22 11:21 am
x = 2
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