1) x^2+2x-35=0
2) x^2-4x-2=0
2006-11-07 11:15:23 · 6 answers · asked by Lion 1 in Science & Mathematics ➔ Mathematics
x^2+2x-35+39=39
x^2+2x+4=39
(x+2)^2 = 39
x+2 = sqrt 39
x = (sqrt 39) - 2
x^2-4x-2=0
x^2-4x-2+6=6
x^2-4x+4=6
(x-2)^2 = 6
x-2 = sqrt 6
x = (sqrt 6) + 2
2006-11-07 11:19:30 · answer #1 · answered by bourqueno77 4 · 0⤊ 1⤋
First get the "number" term on the right side.
You will then take the "middle" coefficient, divide that by 2, and add it to both sides.
Then you can factor easily.
1)
x^2+2x-35=0 , add 35
x^2+2x = 35
(1/2) * 2 = 1 ,, 1^2 = 1 so
x^2+2x +1 = 35 +1
x^2 + 2x + 1 = 36
You can now factor,
(x+1)^2 = 36
solve for x, there will be 2 answers.
x + 1 = 6
x = 5
x + 1 = -6
x = -7
x = -7, or 5
2)
same process as above.
x^2-4x-2=0
x^2-4x = 2
-4/2 = -2 , (-2)^2 = 4
x^2 - 4x + 4 = 2 + 4
x^2 - 4x + 4 = 6
(x-2)^2 = 6
solve for x, there will be 2 answers.
x - 2 = sqrt(6)
x = sqrt(6) + 2
x - 2 = -sqrt(6)
x = -sqrt(6) + 2
x = sqrt(6) + 2 or -sqrt(6) + 2
2006-11-07 19:18:57 · answer #2 · answered by polloloco.rb67 4 · 0⤊ 0⤋
1) x^2+2x-35
Factor:
(x +7)(x-5)
Set each to 0:
x+7=0 or x-5=0
Solve for x:
x= -7 or x=5
2) x^2-4x-2=0
To be honest I'm not sure about this one because the only factors of are 1 and 2 and they can't make four which would make it already factored.
Hope the first one helps though!
2006-11-07 19:21:56 · answer #3 · answered by horslover10 2 · 0⤊ 0⤋
In the first example you could factor the trinomial and set the factors equal to zero. You would not normally complete the square to solve that equation, although you could. In the second example, the expression does not factor, so you would complete the square. Add two to both sides, since the coefficient of the squared term is already one, you just take half the coefficient of the linear term and square it and add that constant to both sides. Then factor the perfect square trinomial on the left; take the square root of both sides of the equation; solve for x. These examples work out nicely, in that the coefficient of the squared term is one. When it isn't, you have to divide through by that coefficient to force it to be one before you can complete the square.
2006-11-07 19:39:57 · answer #4 · answered by anr 3 · 0⤊ 0⤋
ok so
1) x^2 + 2x - 35 = 0
(x + 7)(x - 5) = 0
x = -7, 5
2) x^2 - 4x - 2 = 0
x^2 - 4x + 4 = 6
(x - 2)(x - 2) = 6
(x - 2) = +/- sq. root(6)
x = 2 +/- sq. root(6)
2006-11-07 19:20:34 · answer #5 · answered by deerdanceofdoom 2 · 0⤊ 0⤋
1.x^2+2x=35
adding1
x^2+2x+1=36
(x+1)^2=6^2
x+1=+/-6
x=-1+/-6x=-7 or +5
2.x^2-4x=2
adding 4
x^2-4x+4=6
(x-2)^2=(rt6)^2
x-2=+/-rt6
x=2+/-rt6
x=2+rt6 or 2-rt6
2006-11-07 19:20:29 · answer #6 · answered by raj 7 · 0⤊ 0⤋