Question

underoot 7 y^2 - 6y -13th underoot 7=0

Answer 1

$given \\ \sqrt{7} {y}^{2} - 6y - 13 \sqrt{7 } = 0 \\ \sqrt{7 } {y}^{2} - 13y + 7y - 13 \sqrt{7} = 0 \\ y( \sqrt{7} y - 13) + \sqrt{7} ( \sqrt{7} y - 13) = 0 \\ ( \sqrt{7} y - 13) \: (y + \sqrt{7} ) = 0 \\ ( \sqrt{7} y - 13) = 0 \: and \: (y + \sqrt{7} ) = 0 \\ \sqrt{7} y = 13 \: and \: y = - \sqrt{7} \\ y = \frac{13}{ \sqrt{7} } and \: y = - \sqrt{7} \\ rationalize \: y = \frac{13}{ \sqrt{7} } \\ y = \frac{13 \sqrt{7} }{7} \: and \: y = - \sqrt{7} \\ therefore \: the \: values \: of \: y \: is \\ \frac{13 \sqrt{7} }{7} and \: - \sqrt{7}$