Solve by using the quadratic formula. x^2 - 2x - 15 = 0
2006-11-19 04:53:38 · 13 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
x=[2+/-rt(4+60)]/2
=[2+/-8]/2
=-3 or 5
so the factors are
(x+3)(x-5)
2006-11-19 04:59:33 · answer #1 · answered by raj 7 · 1⤊ 0⤋
AX^2+BX+C=0
so
A=1, B=2, C=-15
shove those into the quadratic formula:
-2+-(SQRT(2^2-4*1*-15))/(2*1)
gives 3 or -5 according to wether you use the + or -
so to factorise,
(X+3)(X-5)=0, so you can see that one of the brackets must be 0
and x is either 3 or -5.
2006-11-19 05:02:08 · answer #2 · answered by jj 2 · 0⤊ 0⤋
x= -5 and 3
2006-11-19 04:59:05 · answer #3 · answered by Casey 3 · 0⤊ 0⤋
5 and -3.
2006-11-19 04:56:28 · answer #4 · answered by Sangmo 5 · 0⤊ 0⤋
2x^2-12x+15 =0 X= 3 (plus or minus) 2 radical 6. X^2+2x-1 =0 X= -1 (plus or minus) radical 2
2016-03-29 01:39:56 · answer #5 · answered by Anonymous · 0⤊ 0⤋
a=1
b= -2
c= -15
Iv'e put a link below so that you can see the quadratic formula, just plug in the numbers.
2006-11-19 05:02:34 · answer #6 · answered by Swirlgirl 3 · 0⤊ 0⤋
If a equation is ax^2+bx+c=0
Then tofind the value of x we can use this formula-------
X= {-b+/-(b^2-4ac)^1/2}/2a
2006-11-19 04:59:40 · answer #7 · answered by sanu 2 · 0⤊ 0⤋
x 1=( 2 + (4+60)^0.5)/2 = 5; X2 =( 2 - (4+60)^0.5)/2= -3
2006-11-19 05:01:35 · answer #8 · answered by maussy 7 · 0⤊ 0⤋
a=1 b=-2 c=-15
-b +- srrt(b2-4ac)
-------------------
2a
= +2 +- sqrt(4-4*1*-15))/2
=( 2+- 8)/2
= 1+-4 or 5 and -3
2006-11-19 05:00:02 · answer #9 · answered by af12af3af 2 · 0⤊ 0⤋
{5, -3}
2006-11-19 04:59:22 · answer #10 · answered by abcde12345 4 · 0⤊ 0⤋