Problema Solution
I NEED THREE DIFFERENT METHODS OF SOLVING QUADRATIC EQUATIONS AND I NEED THE EASIEST AND THE MOST DIFFICULT . I ALSO NEED AN EXAMPLE OF A QUADRATIC EQWUATION WITH TWO DIFFERENT METHODS OF TO BACK UP MY ARGUMENT.. I WOULD ALSO LIKE A REAL WORLD APPLICATION OF A QUADRATIC EQUATION ANDSHOW THE EQUATION AND EXPLAIN WHAT THE VARIABLE REPRESENTS IN REAL TERMS. DESCRIBE WHAT EQUATIONB MODELS IN REAL TERMS.. DO YOU HAVE SOMEONE THERE OR SOFTWARE THAT CAN ANWSER OR DISCUSS WHAT I JUST PROPOSED CONCERNING QUADRATIC EQUAtions???
Answer provided by our tutors
THERE ARE 3 BASIC METHOD TO SOLVE QUADRATIC EQUATIONS---
1) GENERAL METHOD------(or COMPLETING SQUARE METHOD)
ax2+bx+c=0
=> x2+(b/a)x+c/a=0 [dividing both sides of the equation by a not equal to 0)
=> x2+2*x*(b/2a) +(b/2a)2- (b/2a)2+c/a=0
=> (x+b/2a)2= (b/2a)2-c/a
=> (x+b/2a)={(b/2a)2-c/a}0.5 or -{(b/2a)2-c/a}0.5
=> x= -b/2a+ {(b/2a)2-c/a}0.5 or -b/2a- {(b/2a)2-c/a}0.5
2) Sreedhar acharya method or Hindu Method
ax2+bx+c=0
multiply both sides by 4a
4a2x2+4ab+4ac=0
4a2x2+2*(2a)*b + b2-b2+4ac=0
(2ax+b)2=b2-4ac
(2ax+b)= (b2-4ac)0.5 or -(b2-4ac)0.5
2ax=-b +(b2-4ac)0.5 or -b-(b2-4ac)0.5
x=(1/2a)*( -b +(b2-4ac)0.5) or (1/2a)*( -b -(b2-4ac)0.5)
3) Factorisation :-
ax2+bx+c=0
FACTORISE BY THE MIDDLE TERM
AND THEN FIND BOTH THE VALUES OF x