Question

Solve the quadratic equation 5×^2-6×-2 =0 0 by completing the square method​

Answer 1

Hey Pretty Stranger!

$\sf \: 5 {x}^{2} - 6x - 2 = 0$

Dividing by 5

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

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

We know that,

★ ( a - b)² = a² - 2ab + b²

where a = x and

$\sf \: - 2ab = \frac{ - 6x}{5}$

Put the value of a = x

$\sf \: - 2xb = \frac{ - 6x}{5}$

$\sf \: - 2b = \dfrac{ - 6}{5}$

$\sf \: b = \dfrac{ 3}{5}$

Now ,

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

$\sf \: {x}^{2} - \dfrac{6x}{5} - \dfrac{2}{5} + ( \dfrac{3}{5} )^{2} - ( \dfrac{3}{5} ) ^{2} = 0$

$\sf \: {x}^{2} - \dfrac{6x}{5} + ( \dfrac{3}{5} )^{2} - \dfrac{2}{5} - ( \dfrac{3}{5} ) ^{2} = 0$

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

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

$\sf \: (x - \dfrac{3} {5} ) ^{2} = \dfrac{9}{25} + \dfrac{2}{5}$

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

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

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

i) Solving x - 3/5 = + √19/5

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

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

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

ii) Solving x - 3/5 = - √19/5

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

$\sf \: x = - \dfrac{ \sqrt{19} }{5} + \dfrac{3}{5}$

$\sf \: x = \dfrac{ - \sqrt{19 } + 3}{5}$

So,

$\sf \: x = \dfrac{ \sqrt{19 } + 3}{5} \: and \: \dfrac{ - \sqrt{19 } + 3}{5}$

are the roots of the equation.