Question

Solve the following quadratic equation by completing square method.
5y2 + y is equal 3​

Answer 1

Given :-

$5 {y}^{2} + y = 3$

To find :-

y =?

Solution :-

$5 {y}^{2} + y = 3 \\ \frac{5 {y}^{2} }{5} + \frac{y}{5} = \frac{3}{5} \: \: ... \: dividing \: each \: term \: by \: 5 \\ \\ {y}^{2} + \frac{y}{5} = \frac{3}{5} \\ \\ comparing \: with \: a {x}^{2} + bx + c = 0 \: we \: get \: \\ a = 1 \: \: b = \frac{1}{5} \: \: c = \frac{3}{5} \\ \\ now \\ third \: term \: = {( \frac{1}{2} \times b) }^{2} \\ = {( \frac{1}{2} \times \frac{1}{5} )}^{2} \\ = {( \frac{1}{10} )}^{2} \\ \\ = \frac{1}{100} \\ \\ by \: adding \: \frac{1}{100} to \: given \: equation \: we \: get \: \\ \\ {y}^{2} + \frac{y}{5} + \frac{1}{100} = \frac{3}{5} + \frac{1}{100} \\ {(y + \frac{1}{10}) }^{2} = \frac{300 + 5}{500} \\ {(y + \frac{1}{10}) }^{2} = \frac{305}{500} \\ {(y + \frac{1}{10} )}^{2} = \frac{61}{100} \\ \\ y + \frac{1}{10} = + - \sqrt{ \frac{61}{100} } \\ \\ y + \frac{1}{10} = \sqrt{ \frac{61}{100} } \\ y + \frac{1}{10} = \frac{ \sqrt{61} }{10} \\ \\ y = \frac{ \sqrt{61} }{10} - \frac{1}{10} \\ \\ y = \frac{ - 1 + \sqrt{61} }{10} \\ \\ or \: \\ \\ y = \frac{ - 1 - \sqrt{61} }{10}$

Answer :-

y = - 1 + sqt. 61 / 10 OR y = - 1 - sqt. 61 / 10

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