Question

completing square method
$2 {x}^{2} + 4x - 4 = 0$
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Answer 1

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✺Answer:

♦️ To Find RootS:

• $\large{ \rm{ 2 {x}^{2} + 4x - 4 = 0}}$

♦️Method to be used:

• By using completely square method

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✺Explanation of Method:

You must known to the Finding roots method i.e. Sriddhar Acharya's method. The derivation of Sriddhar Acharya's method uses the method of Completing square method which can be used directly to find the roots of a given quadratic equation.

↪ Refer to the attachment...

The Completely square method is the method in which the equation is formed in a such a way that the Left hand side becomes a perfect square in the form of (x+a)^2.

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✺Solution:

Expressing the equation into a perfect square and solving it accordingly.

First of all, making the leading coefficient 1

Diving 2 from both sides.

$\large{ \rm{ \rightarrow \: \frac{2 {x}^{2} + 4x - 4}{2} = \frac{0}{2}}} \\ \\ \large{ \rm{ \rightarrow \: {x}^{2} + 2x - 2 = 0}}$

Transposing 2 to the RHS

$\large{ \rm{ \rightarrow \: {x}^{2} + 2x = 2}}$

Making a perfect square and solving accordingly

$\large{ \rm{ \rightarrow \: (x) {}^{2} + 2 \times x \times 1 + (1) {}^{2} = 2 + (1) {}^{2}}} \\ \\ \large{ \rm{ \rightarrow \: (x + 1) {}^{2} = 2 + 1}} \\ \\ \large{ \rm{ \rightarrow \: x + 1 = \pm \sqrt{3}}} \\ \\ \large{\ddag\: \:{ \boxed{ \bf{ \pink{Case - 1}}}}} \\ \\ \large{ \rm{ \rightarrow \: x + 1 = \sqrt{3} }} \\ \\ \large{ \rm{ \rightarrow \: x = \sqrt{3} - 1}} \\ \\ \large{\ddag\: \:{ \boxed{ \bf{ \pink{Case - 2}}}}} \\ \\ \large{ \rm{ \rightarrow \: x + 1 = - \sqrt{3} }} \\ \\ \large{ \rm{ \rightarrow \: x = - \sqrt{3} - 1}}$

So, the roots of this quadratic equation:

$\large{ \ddag \: \: { \rm{ \underline{ \blue{x = \sqrt{3} - 1 \: or \: - \sqrt{3} - 1}}}}}$

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✺In the attachment :

• Derivation of Sriddhar Acharya's method

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Answer 2

Given ,

The quadratic eq is $\sf 2 {x}^{2} + 4x - 4 = 0$

Divide the given equation by 2

$\sf \Rightarrow {(x)}^{2} - 2x - 2 = 0$

Transpose the constant term to right hand side of the equation

$\sf \Rightarrow {(x)}^{2} - 2x = 2$

Add (b/2)² i.e 1 to both sides of the equation and complete the square

$\sf \Rightarrow {(x)}^{2} - 2x + 1 = 2 + 1 \\ \\ \sf \Rightarrow </p><p> {(x)}^{2} - 2x + {(1)}^{2} = 3 \\ \\ \sf \Rightarrow </p><p> {(x - 1)}^{2} = 3$

Take the square root on both sides of the equation

$\sf \Rightarrow x - 1 = \sqrt{3} \\ \\ \sf \Rightarrow </p><p>x = 1 ± \sqrt{3}$

$\therefore \sf \bold{ \underline{ The \: roots \: of \: quadratic \: equation \: are \: 1 + \sqrt{3} \: and \: 1 - \sqrt{3} \: } }$