Question

By completing sq methods​

Answer 1

$\underline{\textbf{\large{Step-by-step explanation:}}}$

▶finding the roots of quadratic polynomial by completing square method

6x² + x - 2 = 0

Divide by 6 on both the sides

→ $x^2+ \frac {x}{6}- \frac{1}{3} = 0$

→ $x^2+\frac{x}{6}= \frac{1}{3}$

→ $x^2+\frac{1}{6}x = \frac{1}{3}$

Adding both sides( $\frac{1}{2}\times$ (coefficient of x)) =

= $(\frac{1}{2}\times \frac{1}{6})^2$

= $(\frac{1}{12})^2$

=$\frac{1}{144}$

→ $x^2+ \frac{1}{6}x +\frac{1}{144} =\frac{1}{3}+\frac{1}{144}$

→ $(x + (\frac{1}{12}))^2=\frac{1}{3}\times\frac{48}{48} + \frac{1}{144}$

→ $(x + (\frac{1}{12}))^2 = \frac{48}{144} + \frac{1}{144}$

→ $(x + (\frac{1}{12}))^2=\frac{(48 + 1)}{144}$

→ $(x + (\frac{1}{12}))^2= \frac{49}{144}$

→ $(x + (\frac{1}{12}))= \sqrt{(\frac{49}{ 144})}$

→ $(x +\frac{1}{12}) = ± \frac{7}{12}$

→ $x =\frac{7}{12}±\frac{1}{12}$

∴ $x = \frac{(7-1)}{12} = \frac{6}{12} = \frac{1}{2}$

$\boxed{x=\frac{1}{2}}$

OR

∴ $x = \frac{(7+1)}{12}$

$x = \frac{8}{12} = \frac{2}{3}$

$\boxed{x=\frac{2}{3}}$