Question

Find the roots of the given quadratic equation using completing square method
Equation:
${5x}^{2} - 7x - 6 = 0$
​

Answer 1

Step-by-step explanation:

$5 {x}^{2} - 7x - 6 = 0 \\ \\ dividing \: 5 \: on \: both \: sides \\ \\ \frac{5 {x}^{2} }{5} - \frac{7x}{5} - \frac{6}{5} = 0 \\ \\ {x}^{2} - \frac{7x}{5} - \frac{6}{5} = 0 \\ \\ {x}^{2} - \frac{7x}{5} = \frac{6}{5} \\ \\ {x}^{2} - 2 \times \frac{1}{2} \times \frac{7x}{5} = \frac{6}{5} \\ \\ {x}^{2} - 2x \times \frac{7}{10} = \frac{6}{5} \\ \\ adding \: {( \frac{7}{10} )}^{2} \: on \: both \: sides \: \\ \\ {x}^{2} - 2x \times \frac{7}{10} + {( \frac{7}{10} )}^{2} = \frac{6}{5} + { (\frac{7}{10}) }^{2} \\ \\ {(x + \frac{7}{10} )}^{2} = \frac{6}{5} + \frac{49}{100} \\ \\ {( x + \frac{7}{10} )}^{2} = \frac{120 + 49}{100} \\ \\ {(x + \frac{7}{10}) }^{2} = \frac{169}{100} \\ \\ (x + \frac{7}{10}) = \sqrt{ \frac{169}{100} } \\ \\ x + \frac{7}{10} = \sqrt{ \frac{13 \times 13}{10 \times 10} } \\ \\ x + \frac{7}{10} = + - \frac {13}{10} \\ \\ on \: taking \: + \\ \\ x + \frac{7}{10} = \frac{13}{10} \\ \\ x = \frac{13}{10} - \frac{7}{10} \\ \\ x = \frac{13 - 7}{10} \\ \\ x = \frac{6}{10} \\ \\ x = \frac{3}{5} \\ \\on \: taking \: - \\ \\ x + \frac{7}{10} = - \frac{13}{10} \\ \\ x = - \frac{13}{10} - \frac{7}{10} \\ \\ x = = \frac{ - 13 - 7}{10 } \\ \\ x = \frac{ - 20}{10} \\ \\ x = - 2 \\ \\ therefore \: the \: roots \: are \: ( - 2) \: and \: \frac{3}{5}$