Question

Find the roots 9x^2+14x-8=0 competing the square method​

Answer 1

$\huge{\underline{\underline{\green{\mathfrak{Answer:}}}}}$

P(x) = 9x² + 14x - 8

By completing the square method:

$\mathsf {x^2 + \frac{14}{9}x - \frac{8}{9} = 0 }$

$\mathsf {(x + \frac{14}{18})^2 - (\frac{14}{18})^2 - \frac{8}{9} = 0}$

$\mathsf {(x + \frac{14}{18})^2 - (\frac{14}{18})^2 - \frac{8}{9} = 0}$

$\mathsf {(x + \frac{14}{18})^2 - \frac{196}{324} - \frac{8}{9} = 0}$

$\mathsf {(x + \frac{14}{18})^2 - (\frac{196 + 288}{324}) = 0}$

$\mathsf {(x + \frac{14}{18})^2 - \frac{484}{324} = 0}$

$\mathsf {(x + \frac{14}{18})^2 = \frac{484}{324}}$

$\mathsf {x + \frac{14}{18} = ± \sqrt{\frac{484}{324}}}$

$\mathsf {x + \frac{14}{18} = ± \frac{22}{18}}$

$\mathsf {x = \frac{-14}{18} ± \frac{22}{18}}$

$\mathsf {x = \frac{-14}{18} + \frac{22}{18} \:\:\:\:\:\:OR\:\:\:\:\frac{-14}{18} - \frac{22}{18}}$

$\mathsf {x = \frac{8}{18} \:\:\:\:\:\:OR\:\:\:\:\: \frac{-36}{18}}$

$\fbox{\mathsf {x = \frac{4}{9} \:\:\:\:\:\:OR\:\:\:\:\: x = -2}}$

Answer 2

Answer:

This is the written answer for your question.