i finished like 50 of the problems but here's the star problem 10x^2-61x+6=0 and i thank you all so so much u jst do not know
2007-04-12 08:50:26 · 12 answers · asked by Iron Man 1 in Science & Mathematics ➔ Mathematics
10x² - 61x + 6 = 0
The Middle term is - 61x
Find the sum of the middle term
Multiply the first term 10 times the last term 6 equals 60 and factor
The factors of 60 =
1 x 60. . .<=. .use these factors
2 x 30
3 x 20
4 x 15
5 x 12
6 x 10
- 1 and - 60 satisfy the sum of the middle term
- 1 = - x
insert - x and - 60x into the equation
10x² - 61x + 6 = 0
10x² - x - 60x + 6 = 0
x(10x - 1) - 6(10x - 1) = 0
(x - 6)(10x - 1)
- - - - - - - - - - -s-
2007-04-12 09:54:06 · answer #1 · answered by SAMUEL D 7 · 0⤊ 0⤋
just factor it
10x^2 - 61x + 6 = 0
now find two numbers that multiply to equal 60 and add to be -61 (-60 and -1)
so substitute those two numbers with an x for the middle term:
10x^2 - 60x - x + 6 = 0
now you can easily factor the left and right parts:
(10x^2 - 60x) + (-x + 6)
10x(x - 6) - 1(x - 6)
now since the numbers inside the parenthesis are the same, you can multiply them by the numbers outside the parentheses:
(x - 6)(10x - 1) = 0
remember the whole thing is equal to 0 and when two things are multiply together to equal 0, one part or the other has to equal 0, so there will be 2 answers. set each part equal to 0 to get both:
x - 6 = 0.................10x - 1 = 0
x = 6.......................10x = 1
................................x = 1/10
x = 6, 1/10
2007-04-12 08:54:11 · answer #2 · answered by Link 4 · 0⤊ 0⤋
10x² - 61x + 6 = 0
10x² - x - 60x + 6 = 0
x(10x-1) - 6(10x - 1) = 0
(x-6)(10x-1) = 0
x = 6, x = 1/10
2007-04-12 08:54:03 · answer #3 · answered by Anonymous · 2⤊ 0⤋
10x^2-61x+6=0
(10x-1)(x-6)=0
10x-1=0 or x-6=0
x=1/10 or x=6
2007-04-12 08:54:53 · answer #4 · answered by jrome 2 · 2⤊ 0⤋
(10x - 1)(x - 6) = 0
x = 1/10 or x = 6.
To get a large number (61) for the coefficient of x, the numbers multiplying together to add up to the x term need to be large. That is why I tried (10, 1) for the coefficient of x^2 and (1,6) for the absolute terms first, rather than something like (5,2) and (2,3).
2007-04-12 09:01:22 · answer #5 · answered by Anonymous · 0⤊ 0⤋
(10x+1)(x+6) = 0
x= -1/10 OR x= -6
2007-04-12 08:53:40 · answer #6 · answered by MathMark 3 · 0⤊ 1⤋
(10x-1)(x-6)=0
x=1/10; 6
2007-04-12 08:58:41 · answer #7 · answered by Dave aka Spider Monkey 7 · 0⤊ 0⤋
10x^2-60x-x+6=0
10x(x-6)-1(x-6)
x=6 or x =1/10
2007-04-12 08:53:38 · answer #8 · answered by Anonymous · 0⤊ 1⤋
(10x-1)(x-6)=0
x=.1, 6
2007-04-12 08:56:16 · answer #9 · answered by Anonymous · 1⤊ 0⤋
well... i'll try it out its easier than you think
(10x - 1)(x-6)
so when you foil it
10x^2 -60x-x+6
tada
2007-04-12 09:01:28 · answer #10 · answered by c_eckdhal71487 2 · 0⤊ 0⤋