Solve the folowing equation by factoring: x^2-5x-14=0
2006-11-17 15:27:01 · 6 answers · asked by Eric 2 in Education & Reference ➔ Homework Help
x^2-5x-14=0
=>x^2-7x+2x-14=0
=>x(x-7)+2(x-7)=0
=>(x-7)(x+2)=0
Therefore either x-7=0 or x+2=0
If x-7=0,then x=7,and if x+2=0,then x=_2
Thereforex=7 or -2 ans
2006-11-17 15:42:43 · answer #1 · answered by alpha 7 · 0⤊ 0⤋
You are trying to find two numbers that add or subtract to get -5 and two numbers that multiple to get -14 so
-7 and +2 =
-7 x 2 = 14
-7+2=5
so then:
(x-7)(x+2)= make positive negative and vise versa
7,-2
2006-11-18 01:36:42 · answer #2 · answered by dandanthecranman 3 · 0⤊ 0⤋
x^2 - 5x - 14 = 0
by factoring
(x - 7) (x +2) = 0 multiples of 14 to equal -5 are (-7) and (2)
x - 7 = 0...................x + 2 = 0
x = 7........................x = -2
2006-11-17 23:38:36 · answer #3 · answered by choo_hoo 3 · 0⤊ 0⤋
x^2-5x-14=0
(x-7)(x+2)=0
x=-2, 7
2006-11-17 23:30:56 · answer #4 · answered by Anonymous · 1⤊ 0⤋
its 5 + or - the square root of 53 all divided by 2
2006-11-17 23:32:40 · answer #5 · answered by Apple Pie Eater 4 · 0⤊ 0⤋
All you need to do is the quadratic forumla, it should be in your textbook. I'm not sure if this is right, so double check it:
(quadratic foruma) X=-b*(squareroot)b^2*4*a*c
-----------------------------------
2A
2006-11-17 23:30:29 · answer #6 · answered by Icesage0 2 · 0⤊ 0⤋