SOLUTION: How to show that if X1 and X2 are the two roots of {{{ ax^2+bx+c=0 }}} , then {{{ X1X2=c/a }}} and X1+X2=-b/a , where a is not equal 0 ?

Algebra ->  Equations -> SOLUTION: How to show that if X1 and X2 are the two roots of {{{ ax^2+bx+c=0 }}} , then {{{ X1X2=c/a }}} and X1+X2=-b/a , where a is not equal 0 ?      Log On


   

Question 1058234: How to show that if X1 and X2 are the two roots of +ax%5E2%2Bbx%2Bc=0+ , then +X1X2=c%2Fa+
and X1+X2=-b/a , where a is not equal 0 ?
Found 2 solutions by josmiceli, Boreal:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
(1) +a%2Ax%5E2+%2B+b%2Ax+%2B+c+=+0+
Divide both sides by +a+
+x%5E2+%2B+%28b%2Fa%29%2Ax+%2B+c%2Fa+=+0+
----------------------------
Roots are +x%5B1%5D+, and +x%5B2%5D+

(2)
Comparing (1) and (2),
+x%5B1%5D+%2B+x%5B2%5D+=+-b%2Fa+
+x%5B1%5D%2Ax%5B2%5D+=+c%2Fa+



Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
x1=(1/2a)(-b+ sqrt(b^2-4ac))
x2=(1/2a)(-b-sqrt(b^2-4ac))
Their sum is (1/2a)(-2b)=-b/a
Their product is foiling a difference of squares.
That is (1/4a^2)*{b^2-(b^2-4ac)}=4ac/4a^2=c/a, a NE 0