Question

Show that the product of the roots of a quadratic equation is ‘c/a’.

Answer 1

The roots of the quadratic equation are given by x = [-b ± √(b² - 4ac)] / 2a  

so one root is x = [-b + √(b² - 4ac)] / 2a and the second root is x = [-b - √(b² - 4ac)] / 2a  

The product of the roots = {[-b + √(b² - 4ac)] / 2a} * {[-b - √(b² - 4ac)] / 2a}  

= {[-b + √(b² - 4ac)] * [-b - √(b² - 4ac)]} / 4a²  

The numerator is the difference of two squares in reverse ... (a + b)(a - b) = a² - b² ... so:  

= {b² - (b² - 4ac)} / 4a²  

= {b² - b² + 4ac} / 4a²  

= 4ac / 4a²  

= c/a

Answer 2

Solution :

We know that ,

Standard form of Quadratic equation

ax² + bx + c = 0 ( a ≠ 0 )

i ) dividing each term by ' a ' we get

x² + ( b/a )x + c/a = 9

=> x² + ( b/a )x = -c/a

=> x² + 2. x . ( b/2a ) = -c/a

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

=> ( x + b/2a )² = b²/4a² - c/a

=> ( x + b/2a )² = ( b² - 4ac )/4a²

=> ( x + b/2a ) = ± √[(b²-4ac)/(4a²)]

=> x = -b/2a ± ( √b²-4ac )/2a

x = [ -b± ( √b²- 4ac )]/2a

Therefore ,

Two roots of given Quadratic equation

are

{ [ -b+(√b²-4ac)]/2a } , { [-b-(√b²-4ac)]/2a}

Now ,

product of roots

= {[-b+(√b²-4ac)]/2a}{[-b-(√b²-4ac)]/2a}

= { (- b )² - (√ b²-4ac )² }/( 4a² )

= [ b² - ( b² - 4ac ) ]/( 4a² )

= [ b² - b² + 4ac ]/( 4a² )

= ( 4ac )/( 4a² )

= c/a

Therefore ,

Product of the roots of the

Quadratic equation = c/a

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