2006-12-22 00:12:17 · 12 answers · asked by GINA H 2 in Science & Mathematics ➔ Mathematics
3x^ + 13 x -10 = 0
(3x -2) (x + 5) = 0
3x-2 = 0 or x + 5 = 0
x = 2/3 or x = -5
2006-12-22 00:22:38 · answer #1 · answered by mozac 2 · 2⤊ 0⤋
3x^2 + 13x - 10 = 0
*Factor by finding a number divisible by the coefficients >>>>>(3, 13, -10). In this case, there is no divisible number. Next, take the first coefficient (3) and multiply it by the last coefficient (-10) which is -30. Find two numbers that give you -30 when multiplied together and 13 (middle number) when added/subtracted.
*The numbers are: 15, -2
First: rewrite the equation with the new numbers (middle numbers) > group "like" terms:
3x^2 + 15x - 2x - 10 = 0
(3x^2 + 15x) - (2x - 10) = 0
Second: factor each set:
3x(x + 5) - 2(x + 5) = 0
(x + 5)(3x - 2) = 0
Third: solve for each "x" variable > set each parenthesis to equal zero:
1. x + 5 = 0
x + 5 - 5 = 0 - 5
x = -5
2. 3x - 2 = 0
3x - 2 + 2 = 0 + 2
3x = 2
3x/3 = 2/3
x = 2/3
2006-12-22 11:40:49 · answer #2 · answered by ♪♥Annie♥♪ 6 · 2⤊ 0⤋
Here is the traditional way:
1) find two integers that multiply together to make 3 and two more integers that multiply together to make -10. For 3, there are only two choices: (1 and 3) and (-1 and -3), For -10, there are four choices: (1, -10), (-1, 10), (-2, 5) and (2, -5).
2) Pick one set from each, say (1, 3) and (-2, 5)
3) Try to cross multiply two by two so that the sum of products gives you the middle factor +13.
In this case, we can use (3 times 5) plus (1 times -2) = 15 -2 = 13.
4) Then the factors are given by switching (i.e., taking the 3 with the -2, and the 1 with the 5).
(3x - 2)(x +5)
In this example, I used the correct ones the first time around. Sometimes you have to try a few combinations before hitting the right one.
Also, the symmetrical choices could have worked (-1, -3) and (2, -5): (-3x +2)(-x - 5) = 3x^2 +13x -10
2006-12-22 08:36:33 · answer #3 · answered by Raymond 7 · 2⤊ 0⤋
3x^2 + 13x - 10 = 0
<=> 3x^2 + 15x - 2x - 10 =0
<=>3x(x + 5) - 2(x+5) = 0
<=>(3x -2)(x + 5) = 0
<=> x = 2/3 or x = -5
2006-12-22 08:25:19 · answer #4 · answered by James Chan 4 · 3⤊ 0⤋
(3x - 2)(x + 5) = 0
3x-2 = 0 or x + 5 = 0
3x = 2 x = -5
x = 2/3
So the roots of 3x^2 + 13x - 10 = y are at x = -5 and x = 2/3
2006-12-22 08:29:12 · answer #5 · answered by stewartlucas467 2 · 2⤊ 0⤋
3x^2 + 13x - 10 = 0
3x^2 + (15 - 2)x - 10 = 0
3x ( x + 5) - 2 (x + 5) = 0
(3x - 2) (x + 5) = 0
x = -5, 2/3
2006-12-22 09:59:53 · answer #6 · answered by Anonymous · 3⤊ 0⤋
3x^2+13x-10==0
=>3x^2+15x-2x-10=0
=>3x[x+5]-2[x+5]=0
=>[3x-2][x+5]=0
=>x=2/3 or -5
2006-12-22 09:37:50 · answer #7 · answered by Optimist!*$/905040 1 · 3⤊ 0⤋
i think ur q must be
A.3x^2-13x-10=0
or it may be
B.3x^2+13x+10=0
check ur question again.
If it is question in A, then
3x^2-10x-3x-10=0
x(3x-10)-1(3x-10)=0
(x-1)(3x-10)=0
so,
x=1 or
x=10/3
If it is the question in B then put + in all the places instead of minus...and u will get
(x+1)(3x+10)=0
so
x= -1
or
x= -10/3
2006-12-22 08:27:57 · answer #8 · answered by sri_july27 2 · 0⤊ 4⤋
3x^2+13x-10=0
[-13+/-root(13^2-4*3*-10)]/2*3
x1= 2/3
x2=-5
2006-12-22 08:29:07 · answer #9 · answered by Anonymous · 2⤊ 0⤋
3x^2+13x-10=0
3x^2-15x+2x-10=0
3x[x-5]+2[x-5]=0
[x-5][3x+2]=0
x=5,-2/3
2006-12-22 10:52:40 · answer #10 · answered by openpsychy 6 · 3⤊ 0⤋