how do i do it???
plzz help mee
tanx
2007-10-17 10:22:28 · 13 answers · asked by -x-lollie-x- 1 in Science & Mathematics ➔ Mathematics
5x^2-7x-6
Multiply the coefficient of x^2 which is 5 with the last term which is 6 to get 30, then think of two numbers whose sum will give you the coefficient of the middle term (-7) which is
-10 and 3. In this case, -10x+3x.
The equation then becomes, 5x^2-10x+3x-6
5x^2-10x+3x-6
(5x^2-10x) +(3x-6)
Factor out 5x from the 1st bracket, and 3 from 2nd bracket
5x(x-2)+3(x-2)
(x-2)(5x+3)
Finding the value of x, equate the above to zero
(x-2)(5x+3)=0
x-2=0, x=2
5x+3=0, x= -3/5
2007-10-17 22:39:10 · answer #1 · answered by Ama 3 · 0⤊ 0⤋
5x^2 - 7x - 6 = 0
(5x + 3)(x-2) = 0
set each one of the two terms = 0
5x + 3 = 0
5x = -3
x = -3/5
or
x - 2 =0
x = 2
check answers in original equation
for x = -3/5
5x^2 - 7x - 6
5(-3/5)^2 - 7(-3/5) - 6
9/5 + 21/5 - 6
30/5 - 6
6 - 6 = 0 checks out
for x = 2
5x^2 - 7x - 6
5(2)^2 -7(2) - 6
20 - 14 - 6 = 0 checks out
2007-10-17 10:29:45 · answer #2 · answered by tiger 2 · 0⤊ 0⤋
here u go
Simplifying
5x2 + -7x + -6 = 0
Reorder the terms:
-6 + -7x + 5x2 = 0
Solving
-6 + -7x + 5x2 = 0
Solving for variable 'x'.
Factor a trinomial.
(-3 + -5x)(2 + -1x) = 0
2007-10-17 10:27:43 · answer #3 · answered by phil m 2 · 0⤊ 0⤋
5x^2-7x-6
factor out
(5x+3)(x-2).
to find the value of x
(5x+3)=0 or(x-2)=0
5x=-3 or x-2=0
x=-3/5 or x=2
2007-10-17 14:41:50 · answer #4 · answered by tapsmat1 2 · 0⤊ 0⤋
I can factor it.
5x^2-7x-6=(5x+3)(x-2)
If you want it solved, then we must say
(5x+3)(x-2)=0
Either(5x+3)=0 or (x-2)=0
If 5x+3=0, 5x=-3, x=-3/5
Ifx-2=0, x=2
2007-10-17 10:29:45 · answer #5 · answered by Grampedo 7 · 0⤊ 0⤋
5x^2 - 7x - 6 = 0
( 5x + 3) (x - 2) = 0
x = -3/5
or x = 2
2007-10-18 01:04:48 · answer #6 · answered by Nick H 2 · 0⤊ 0⤋
Tell to what value it should be solved...
I guess it might be for zero.
Then the answers will be
(-(-7) + sqrt(7^2 - 4 * 5 * (-6)))/(2 * 5)
and
(-(-7) - sqrt(7^2 - 4 * 5 * (-6)))/(2 * 5)
=> 2 and -3/5 respectively!
2007-10-17 10:31:18 · answer #7 · answered by Karteek :: The K' Factor 2 · 0⤊ 0⤋
There is not an = sign so cannot be solved.
It can be factorised:-
( 5 x + 3 ) ( x - 2 )
If indeed there is an = sign:-
x = - 3 / 5 , x = 2 is solution
2007-10-17 20:01:09 · answer #8 · answered by Como 7 · 0⤊ 0⤋
use the quadratic formula:
[-b +/- sqrt (b^2 -4ac) ] /2a
[-(-7) +/- sqrt ((-7)^2 -4(5)(-6)) ] /2(5)
[7 +/- sqrt (49 +120) ] /10
[7 +/- sqrt (169) ] /10
[7 +/- 13 ] /10
(7+13)/10 = 20 /10 = 2
(7-13) /10 = -6/10 = -3/5
(x-2) (5x+3)
2007-10-17 10:29:14 · answer #9 · answered by sayamiam 6 · 0⤊ 0⤋
(5x+3)(x-2) = 0
x = -3/5, 2
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Ideas: Factor 5x^2 * (-6) into (-10x)*(3x), since -10x + 3x = -7x.
2007-10-17 10:28:01 · answer #10 · answered by sahsjing 7 · 0⤊ 0⤋