Question

Find the roots
ef
Eq by the method of
completing the square 5x² - 6x - 2 = 0​

Answer 1

Given :

5x² - 6x - 2 = 0

To find :

Root of above equation by completing Square method

Solution :

1) First of all , make coefficient of x² a perfect square.

Here we are given that coefficient of x² is 5 .

So we will take 5 as common from whole Equation.

$\sf \implies \: 5( {x}^{2} \: - \: \dfrac{6}{5} x \: - \dfrac{2}{5} \: )= \: 0$

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2) We can write above equation as ,

$\sf \implies \: 5( {x}^{2} \: + \: 2 \times ( x) \times( \dfrac{- 3}{5} ) \: - \dfrac{2}{5} \: )= \: 0$

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3) Add and subtract (-3/5)² on LHS { inside the bracket }

$\sf \implies \: 5 \left[ \: \: {x}^{2} \: + \: 2 \times ( x) \times( \dfrac{- 3}{5} ) \: - \dfrac{2}{5} \: + ( { \dfrac{ - 3}{5} }^{2} ) - ( { \dfrac{ - 3}{5} }^{2}) \: \: \right] \: = \: 0 \\ \\ \\ \sf \implies \: \left[ \: \bigg \{{x}^{2} \: + \: 2 \times ( x) \times( \dfrac{- 3}{5} ) \: + ( { \dfrac{ - 3}{5} }^{2} ) \bigg \} - \dfrac{2}{5} \: - ( { \dfrac{ - 3}{5} }^{2}) \: \right] = \: \dfrac{0}{5}\\ \\ \\ \sf \implies \: \left[ \: \bigg \{{x}^{2} \: + \: 2 \times ( x) \times( \dfrac{- 3}{5} ) \: + ( { \dfrac{ - 3}{5} }^{2} ) \bigg \} - \dfrac{2}{5} \: - ( { \dfrac{ - 3}{5} }^{2}) \: \right] = \: 0$

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4) Compare it with (a+b)² = a² + 2(a)(b) + b²

{ a = x ; b = -(3/5) }

$\sf \implies \: \left[ \: { \bigg \{ x + \dfrac{( - 3)}{5} \bigg \}}^{2} - \dfrac{2}{5} \: - \dfrac{ 9}{25} \: \right] = \: 0 \\ \\ \\ \sf \implies \: \left[ \: { \bigg \{ x - \dfrac{3}{5} \bigg \}}^{2} - \dfrac{(2 \times 5) + (1 \times 9)}{25} \: \right] = \: 0 \\ \\ \\ \sf \implies \: \left[ \: { \bigg \{ x - \dfrac{3}{5} \bigg \}}^{2} - \dfrac{(10) + ( 9)}{25} \: \: \right] = \: 0 \\ \\ \\ \sf \implies \: \left[ \: { \bigg \{ x - \dfrac{ 3}{5} \bigg \}}^{2} - \dfrac{19}{25} \ \: \right] = \: 0 \\ \\ \\ \sf \implies \: { \bigg \{ x - \dfrac{ 3}{5} \bigg \}}^{2} \ \: = \: \dfrac{19}{25}$

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5) Take square root on both side & solve for x

$\sf \implies \sqrt{( {x - \dfrac{3}{5}) }^{2} } \: = \: \sqrt{ \dfrac{19}{25} } \\ \\ \\ \sf \implies (x - \dfrac{3}{5} ) = \pm \: \dfrac{ \sqrt{19} }{5} \\ \\ \\ \sf \implies x \: = \dfrac{3}{5} \pm \: \dfrac{ \sqrt{19} }{5} \\ \\ \\ \sf \implies x \: = \bold{ \dfrac{3 \pm \: \sqrt{19} }{5} }$

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ANSWER :

Root of given equation ➝

$\sf \: \: \: x \: = \bold{ \dfrac{3 \pm \: \sqrt{19} }{5} }$