2 + (sqrt) 2, 2 - (sqrt) 2
2006-11-19 15:39:06 · 10 answers · asked by deano899 1 in Science & Mathematics ➔ Mathematics
so easy :
x^2 - Sx + P = 0
where, S is the sum of the roots and P is the Product of the Roots :
S = 2+(sqrt)2 + 2-(sqrt)2 = 4
P = [2+(sqrt)2][2-(sqrt)2]= 4-(2) = 2
==> the equation is x^2 - 4x + 2 = 0
hope this helps enjoy !!??!!
2006-11-19 16:16:53 · answer #1 · answered by Yogesh G 3 · 1⤊ 0⤋
even as looking the quadratic equation from given numbers, purely subtract them from x (because both area of the equation has to equivalent 0: (x + 7)(x - 4) = x^2 + 7x - 4x - 28 = x^2 + 3x - 28
2016-11-29 07:20:19 · answer #2 · answered by ? 4 · 0⤊ 0⤋
You have two given roots 2 + (sqrt) 2, 2 - (sqrt) 2.
Let r1 = 2 + (sqrt) 2 and r2 = 2 - (sqrt) 2
THen the equation with these two roots is
(x-r1)(x-r2)=0
=>(x-(2 + (sqrt) 2))(x-(2 - (sqrt) 2))=0
=>x*(x-(2 + (sqrt) 2))-(2 - (sqrt) 2)*(x-(2 + (sqrt) 2))=0
=>x*(x-2 - (sqrt) 2)) - (2 - (sqrt) 2)*(x-2 - (sqrt) 2))=0
=>x^2 - 2x - ((sqrt) 2)x - [2*(x-2 - (sqrt) 2) - (sqrt) 2)*(x-2 - (sqrt) 2)=0
=>x^2 - 2x - ((sqrt) 2)x - [2x -4 - ((sqrt) 2)2 - (sqrt) 2)*x +2(sqrt) 2)+ (sqrt) 2)*(sqrt) 2)=0
=>x^2 - 2x - ((sqrt) 2)x - 2x + 4 + ((sqrt) 2)2 + (sqrt) 2)*x -2((sqrt) 2) - (sqrt) 2)*(sqrt) 2)=0
=>x^2 - 2x - 2x + 4 - (sqrt) 2)*(sqrt) 2)=0
=>x^2 - 4x + 4 - 2=0
=>x^2 - 4x + 2=0
This is your required equation :
2006-11-20 01:00:39 · answer #3 · answered by Paritosh Vasava 3 · 0⤊ 0⤋
x^2 - 4x + 2 = 0
This is because if a and b are the roots of a quadratic equation then
(x - a)(x - b) = 0
ie x² - (a + b)x + ab = 0
Now in this case a + b = 4
and ab = (2 + √2)(2 - √2)
= 2² - √2²
= 4 - 2
= 2
2006-11-19 15:41:54 · answer #4 · answered by Wal C 6 · 0⤊ 1⤋
(x-2+(sqrt) 2) (x - 2 - (sqrt) 2) = 0
or (x-2)^2 - ((sqrt) 2)^ 2 = 0
or (x-2)^2 - 2 = 0
= x^2 - 4x + 4 - 2 = 0
= x^2 - 4x +2 = 0
2006-11-19 15:46:49 · answer #5 · answered by spoonish18 2 · 0⤊ 1⤋
This means that these are your zeros, so that means the following are the factors of your equation:
(x - (2+√2))(x - (2-√2))
Multiply out using FOIL to get the following:
x² - (2+√2)x - (2-√2)x + (2+√2)(2-√2) =
x² - 2x - √2x - 2x + √2x + 2 =
x² - 4x + 2
Hope that helped!
~ ♥ ~
2006-11-19 15:42:42 · answer #6 · answered by I ♥ AUG 6 · 0⤊ 1⤋
There is no unique solution. Any quadratic that satisfiies the eqns
8a^2 -4ac = 0 and b = -4a will give the roots you specify.
2006-11-19 15:52:50 · answer #7 · answered by Steve 7 · 1⤊ 1⤋
This is because if a and b are the roots of a quadratic equation then
(x - a)(x - b) = 0
ie x² - (a + b)x + ab = 0
2006-11-19 16:01:33 · answer #8 · answered by arpita 5 · 0⤊ 1⤋
x=2+√2
x-(2+√2)=0
x=2-√2
x-(2-√2)=0
(x-(2+√2))(x-(2-(2-√2))
x^2-4x+2=0
2006-11-19 15:45:25 · answer #9 · answered by yupchagee 7 · 0⤊ 1⤋
(x-2-rt2)(x-2+rt2)
x^2-2x+rt2x-2x+4-2rt2-rt2x+2rt2-2
=x^2-4x+2
2006-11-19 16:08:52 · answer #10 · answered by raj 7 · 0⤊ 1⤋