Question

Given that quadratic equation ax²+bx+c=0, then prove that. x= –b±√b²–4ca / 2a.

Answer 1

It can be solved by applying completing the square method.

Answer 2

Given  

a x2 + b x + c = 0  

Divide all terms by a  

x2 + b / a x + c / a = 0  

Subtract c / a from both sides  

x2 + (b / a) x + c / a - c/ a = 0 - c / a  

and simplify  

x2 + (b / a) x = - c / a  

Add (b / 2a)2 to both sides  

x2 + (b / a) x + (b / 2a)2 = - c / a + (b / 2a)2

and complete the square on the left hand side of the equation  

(x + b / 2a )2 = - c / a + (b / 2a)2

Group the two terms on the right side of the equation  

(x + b / 2a)2 = (b2 - 4 a c) / (4 a2 )  

Solve by taking the square root  

x + b / 2a = ~+mn~ √( (b2 - 4 a c ) / (4 a2) )  

Solve for x to obtain two solutions  

x = - b / 2a ~+mn~ √( (b2 - 4 a c ) / (4 a2) )  

The term √( (b2 - 4 a c ) / (4 a2) ) may be simplified as follows  

√( (b2 - 4 a c ) / (4 a2) ) = √(b2 - 4 a c) / (2 |a|)  

Since 2 | a | = 2 a for a > 0 and 2 | a | = - 2 a for a < 0, the two solutions to the quadratic equation may be written  

x1 = ( - b + √( b2 - 4 a c)) / (2 a)  

x2 = ( - b - √ ( b2 - 4 a c)) / (2 a)