Question

Solve the following quadratic equations.
$\red{ \boxed \dag }\: \green{ \tt \: {x}^{2} + 3x - 21 = 0}$
$\large \bold{ \dag }\: \small \: \tt \: prove \: that \: \bold{ \tt \: \orange{x = \dfrac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a} }}$
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Good luck!!​

Answer 1

Topic :-

Quadratic Equation

To Prove :-

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

Proof :-

General form of quadratic equation :

ax² + bx + c = 0 where a ≠ 0.

Divide by 'a' in LHS and RHS,

$\dfrac{ax^2}{a}+\dfrac{bx}{a}+\dfrac{c}{a}=\dfrac{0}{a}$

$x^2+\dfrac{bx}{a}+\dfrac{c}{a}=0$

Add and Subtract (b/2a)²,

$x^2+\dfrac{bx}{a}+\dfrac{c}{a}+\left( \dfrac{b}{2a}\right )^2-\left( \dfrac{b}{2a}\right )^2=0$

Rearrange it,

$x^2+\dfrac{bx}{a}+\left( \dfrac{b}{2a}\right )^2+\dfrac{c}{a}-\left( \dfrac{b}{2a}\right )^2=0$

Write (bx/a) as 2(x)(b/2a),

$x^2+2\dfrac{bx}{2a}+\left( \dfrac{b}{2a}\right )^2+\dfrac{c}{a}-\left( \dfrac{b}{2a}\right )^2=0$

We can observe p² + 2pq + q² = (p + q)² forming here,

$x^2+2(x)\left(\dfrac{b}{2a}\right)+\left( \dfrac{b}{2a}\right )^2+\dfrac{c}{a}-\left( \dfrac{b}{2a}\right )^2=0$

$\left( x+\dfrac{b}{2a} \right )^2+\dfrac{c}{a}-\left( \dfrac{b}{2a}\right )^2=0$

Transposing (c/a) - (b/2a)² to RHS,

$\left( x+\dfrac{b}{2a} \right )^2=\left( \dfrac{b}{2a}\right )^2-\dfrac{c}{a}$

Taking LCM in RHS and solving it,

$\left( x+\dfrac{b}{2a} \right )^2=\dfrac{b^2}{4a^2}-\dfrac{c}{a}$

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

Taking Root both sides,

$\sqrt{\left( x+\dfrac{b}{2a} \right )^2}=\sqrt{\dfrac{b^2-4ac}{4a^2}}$

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

Transpose (b/2a) to RHS,

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

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

Taking LCM in RHS and solving,

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

Hence, Proved !!

--------------------------------------------------------------

Given :-

x² + 3x - 21 = 0

To Find :-

Value of 'x'.

Solution :-

Let us solve using the formula which we have proved above.

On comparing with general equation,

a = 1

b = 3

c = -21

Applying formula,

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

$x=\dfrac{-3 \pm \sqrt{(3)^2-4(1)(-21)}}{2(1)}$

$x=\dfrac{-3 \pm \sqrt{9+84}}{2}$

$x=\dfrac{-3 \pm \sqrt{93}}{2}$

Answer :-

So, value of x are :-

$x=\dfrac{-3 + \sqrt{93}}{2}\:\:and$

$x=\dfrac{-3 - \sqrt{93}}{2}$

Answer 2

Question :

Solve the following quadratic equations.

${ \tt \: {x}^{2} + 3x - 21 = 0}$

Solution :

$\dag \tt \: \: {x}^{2} + 3x - 21 = 0$

✯ Using quadratic formula

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

Here ,

• a = 1
• b = 3
• c = -21

$: \leadsto \sf \: x = \dfrac{ - 3 \pm \sqrt{ {3}^{2} - 4 \times 1 \times - 21 } }{2 \times 1}$

$: \leadsto \sf x = \dfrac{ - 3 \pm \sqrt{ 9 - 84 } }{2}$

$: \leadsto \sf x = \dfrac{ - 3 \pm \sqrt{ 93 } }{2}$

$: \leadsto \sf x = \dfrac{ - 3 \pm 9.64 }{2}$

Taking positive

$: \leadsto \sf x = \dfrac{ - 3 + 9.64 }{2}$

$: \leadsto \sf x = \dfrac{ 6.64 }{2}$

$: \leadsto \sf x = 3.32$

Taking negative

$: \leadsto \sf x = \dfrac{ - 3 - 9.64 }{2}$

$: \leadsto \sf x = \dfrac{ - 12.64 }{2}$

$: \leadsto \sf x = - 6.32$

$\boxed{ \sf \: x = 3.32 \: or \: - 6.32}$

_____________________

Question :

$\large \bold{ \dag }\: \small \: \tt \: prove \: that \: \\ \bold{ \tt \: {x = \dfrac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a} }}$

Solution :

Let the quadratic equation be ax² + bx + c , a≠0

$\hookrightarrow \sf a {x}^{2} + bx + c = 0$

$\hookrightarrow \sf a {x}^{2} + bx = - c \: \: \: \: ( transposing \: the \: constant \: term)$

$\hookrightarrow\sf \: {x}^{2} + \dfrac{b}{a} x = - \dfrac{c}{a} \: \: \: \: (dividing \: by \: cofficient \: of \: {x}^{2} )$

Adding ${ \sf\frac{ {b}^{2} }{4{a}^{2}}}$ to both sides to make LHS as perfect square.

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

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

Taking the square root of both sides

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

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

$\implies { \tt {x}} = \mathfrak{ \dfrac{ - b \: \pm \: \sqrt{ {b}^{2} - 4ac} }{2a} }$

${ \large{ \purple{ \boxed{ \sf \dag \: hence \: proved \dag}}}}$

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