SOLUTION: Prove that the roots of quadratic equation i.e ax2+bx+c=0 numerically equals but opposite in sign to the roots of ax2-bx+c

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Question 1049893: Prove that the roots of quadratic equation i.e ax2+bx+c=0 numerically equals but opposite in sign to the roots of ax2-bx+c
Answer by advanced_Learner(501) About Me  (Show Source):
You can put this solution on YOUR website!
ax%5E2%2Bbx%2Bc=0
x%5E2%2B%28b%2Fa%29x%2Bc%2F%28a%29=0
x%5E2%2B%28b%2Fa%29x=-c%2Fa

x%5E2%2B%28b%2Fa%29x%2B%28b%2F2a%29%5E2=-c%2Fa+%28b%2F2a%29%5E2
%28x%2Bb%2F2a%29%5E2=-c%2Fa+%28b%2F2a%29%5E2

%28x%29=+-b%2F2a+%2B-+sqrt%28-c%2Fa%2B%28b%2F2a%29%5E2%29

x+=+%28-b+%2B-+sqrt%28+-4%2Aa%2Ac%2Bb%5E2+%29%29%2F%282%2Aa%29%29+
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
ax%5E2-bx%2Bc=0

x%5E2-%28b%2Fa%29x%2Bc%2F%28a%29=0
x%5E2-%28b%2Fa%29x=-c%2Fa

x%5E2-%28b%2Fa%29x%2B%28b%2F2a%29%5E2=-c%2Fa+%28b%2F2a%29%5E2
%28x-b%2F2a%29%5E2=-c%2Fa+%28b%2F2a%29%5E2

%28x%29=+b%2F2a+%2B-+sqrt%28-c%2Fa%2B%28b%2F2a%29%5E2%29


x+=+%28b+%2B-+sqrt%28+-4%2Aa%2Ac%2Bb%5E2+%29%29%2F%282%2Aa%29%29+
x+=+%28b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
hence proved