Question

Prove the quadratic formula by completing the square

Answer 1

HEYA FRIEND

Here is ur answer ✍
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QUADRATIC EQUATION 【SHREEDHARACHARYA'S RULE】

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CONSIDER THE QUADRATIC EQUATION ax^2+bx+c.=0, where a, b, c ate real numbers and a✖=0, then .

ax^2+bx+c=0

➡ax^2+bx=-c

➡x^2+b/a.x=-c/a 【dividing throughout by a】


➡x^2+b/a.x+(b/2a)^2=-c/a+(b/2a)^2

【adding b/2a)^3on both sides 】

➡(x+b/2a)^2=(-c/a+b^2/4a^2)

➡(x+b/2a)^2=(b^2-4ac)/4a^2

➡(x+b/2a)^2=+-√b^2-4ac/4a^2

➡(x+b/2a)=+-√b^2-4ac/2a ,

where (b^2-4ac)>=0

=>x=-b/2a+-√b^2-4ac/2a
.

x=-b+-√b^2-4ac/2a....


THIS IS CALLED THE QUADRATIC FORMULA OF SHREEXDHRACHARYA'S RULE

THUS ,ax^2+bx+c=0 has two roots @ and B(bita ), given by

@=-b+-b^2-4ac/2a

and B (bita )=-b-√b^2-4ac/2a....

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Hope it helps you...

@Rajukumar☺☺☺

Answer 2

Hi ,

Let ax² + bx + c = 0 , where a, b , c are

real numbers , a ≠ 0 , is a quadratic

equation ,

Finding roots by completing square

method:
__________

ax² + bx + c = 0 ----( 1 )

Divide each term with 'a ' , we get

x² + ( b/a ) x + c/a = 0

x² + ( b/a )x = ( - c/a )

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

( x + b/2a )² = ( b² - 4ac ) / 4a²

x + b/2a = ± √ [ ( b² - 4ac ) / (2a )² ]

x + b / 2a = ± √ ( b² - 4ac ) / 2a

Therefore ,

x = - ( b/2a ) ± √ ( b² - 4ac ) / 2a

x = [ -b ± √ ( b² - 4ac ) ] / 2a

I hope this helps you.