Question

in this solve any 1 pls​

Answer 1

Given : $\sf{ ax^2 + bx + c = 0}$


⇒ $\sf{ax^2 + bx = - c}$


Divide all the terms with the coefficient of x² i.e. a.


⇒ $\sf{ {\dfrac{ax^2}{a}} + {\dfrac{bx}{a}} = {\dfrac{- c}{a}}}$


⇒ $\sf{ {\dfrac{ {\cancel{a}} x^2}{ {\cancel{a}} }} + {\dfrac{b}{a}} x = {\dfrac{- c}{a}} }$


⇒ $\sf{x^2 + {\dfrac{b}{a}} x = {\dfrac{- c}{a}}}$


Add the square of the half of the coefficient of x on both sides i.e. L.H.S. and R.H.S.


⇒ $\sf{x^2 + {\dfrac{b}{a}} x + {\dfrac{b^2}{4a^2}} = {\dfrac{- c}{a}} + {\dfrac{b^2}{4a^2}} }$


Identity : a² + 2ab + b² = (a + b)²


Here, a= x, b = b/2a


⇒ $\sf{ (x + {\dfrac{b}{2a}})^2 = {\dfrac{- 4ac + b^2}{4a^2}}}$


⇒ $\sf{ (x + {\dfrac{b}{2a}})^2 = {\dfrac{b^2 - 4ac}{4a^2}}}$


⇒ $\sf{x + {\dfrac{b}{2a}} = {\sqrt{ {\dfrac{b^2 - 4ac}{4a^2}} }}}$


⇒ $\sf{x + {\dfrac{b}{2a}} = ± {\dfrac{ {\sqrt{b^2 - 4ac}} }{2a}} }$


⇒ $\sf{x = {\dfrac{- b}{2a}} ± {\dfrac{ {\sqrt{b^2 - 4ac}}}{2a}} }$


⇒ ${\boxed{\sf{x = {\dfrac{- b ± {\sqrt{b^2 - 4ac}} }{2a}} }}}$

Answer 2

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GOOD MORNING!!

ax² + bx + c = 0

DIVIDE BOTH SIDE'S BY a

x² + bx/a + c/a = 0

=>

x² + 2 ( x ) ( b/2a ) + b²/4a² - b²/4a² +c/a = 0

=>

{ x + b/2a }² ={ b² - 4ac }/4a²=0

=>

{ x + b/2a } = ±√{b² - 4ac } / 2a

=>

x = { -b ± √ { b² - 4ac } } /2a