Question

solve by completing square​

Answer 1

$\huge\mathcal{\fcolorbox{cyan}{black}{\pink{ANSWER}}}$

Given,

$a {x}^{2} + bx + c = 0 \\ \implies \: 4 {a}^{2} {x}^{2} + 4abx + 4ac = 0 \: \: \: \{ \: multiply \: by \: 4a \} \\ \implies \: \: \underbrace{ {(2ax)}^{2} + 2 \times 2ax \times b + {b}^{2} } _{ \large \: \: {{(2ax + b)}^{2} }} + 4ac = {b}^{2} \: \: \: \{ \: add \: \: {b}^{2} \: in \: both \: side \} \\ \implies \: {(2ax + b)}^{2} + 4ac = {b }^{2} \\ \implies \: \: {(2ax + b)}^{2} = {b}^{2} - 4ac \\ \implies \: 2ax + b = \pm \sqrt{ {b}^{2} - 4ac} \\ \implies \: 2ax = - b \pm \sqrt{ {b}^{2} - 4ac} \\ \implies \: x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a} \\ \\ \\ \therefore \: x_1 = \frac{ - b + \sqrt{ {b}^{2} - 4ac } }{2a} \\ \\ x_2 = \frac{ - b - \sqrt{ {b}^{2} - 4ac } }{2a}$

Answer 2

$\large\underline{\sf{Solution-}}$

Given quadratic equation is

$\rm :\longmapsto\: {ax}^{2} + bx + c = 0$

Step :- 1 Take out the constant term on RHS

$\rm :\longmapsto\: {ax}^{2} + bx = - c$

Step :- 2 Make the coefficient of x² unity. So Divide whole equation by a.

$\rm :\longmapsto\: {x}^{2} + \dfrac{b}{a}x = - \dfrac{c}{a}$

Step :- 3 Add the square of half the coefficient of x on both sides.

$\rm :\longmapsto\: {x}^{2} + \dfrac{b}{a}x + {\bigg(\dfrac{b}{2a} \bigg) }^{2} = - \dfrac{c}{a} + {\bigg(\dfrac{b}{2a} \bigg) }^{2}$

can be rewritten as

$\rm :\longmapsto\: {x}^{2} + 2 \times \dfrac{b}{2a}x + {\bigg(\dfrac{b}{2a} \bigg) }^{2} = - \dfrac{c}{a} + {\bigg(\dfrac{ {b}^{2} }{ {4a}^{2} } \bigg) }$

We know,

$\: \: \: \: \: \: \: \: \boxed{ \bf{ {x}^{2} + 2xy + {y}^{2} = {(x + y)}^{2}}}$

So using this identity in LHS of above expression, we get

$\rm :\longmapsto\:{\bigg(x + \dfrac{b}{2a} \bigg) }^{2} = {\bigg(\dfrac{ - 4ac + {b}^{2} }{ {4a}^{2} } \bigg) }$

can be rewritten as

$\rm :\longmapsto\:{\bigg(x + \dfrac{b}{2a} \bigg) }^{2} = {\dfrac{{b}^{2} - 4ac }{ {4a}^{2} } }$

$\rm :\longmapsto\:{x + \dfrac{b}{2a}} = \: \pm \: \sqrt{{\dfrac{{b}^{2} - 4ac }{ {4a}^{2} } }}$

$\rm :\longmapsto\:{x + \dfrac{b}{2a}} = \: \pm \: \dfrac{ \sqrt{{b}^{2} - 4ac }}{2a}$

$\rm :\longmapsto\:{x = - \dfrac{b}{2a}} \: \pm \: \dfrac{ \sqrt{{b}^{2} - 4ac }}{2a}$

$\rm :\longmapsto\:{x = } \: \dfrac{ - b \: \pm \: \sqrt{{b}^{2} - 4ac }}{2a}$

Hence,

The solution of

$\bf :\longmapsto\: {ax}^{2} + bx + c = 0$

is

$\bf :\longmapsto\:{x = } \: \dfrac{ - b \: \pm \: \sqrt{{b}^{2} - 4ac }}{2a}$

Additional Information :-

The term b² - 4ac in above expression is called Discriminant and is represented by D and is used in quadratic equations to find the nature of roots.

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

If Discriminant, D > 0, then roots of the equation are real and unequal.

If Discriminant, D = 0, then roots of the equation are real and equal.

If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

• Discriminant, D = b² - 4ac