Find the roots of $- 9 x^{2} - 26 x + 3 = 0$

Open in App

Solution

$- 9 x^{2} - 26 x + 3 = 0$
$9 x^{2} + 26 x - 3 = 0$
$9 x^{2} + 27 x - x - 3 = 0$
$9 x (x + 3) - 1 (x + 3) = 0$
$x = - 3$ or $x = \frac{1}{9}$

Suggest Corrections

Similar questions

tan \frac{α}{2}

tan \frac{β}{2}

9 x^{2} - 26 x + 15 = 0

cos (α + β)

x^{4} - 26 x^{2} + 25 = 0

x^{3} - 9 x^{2} + 26 x - 24

3 x^{2} - 2 \sqrt{6} x + 2 = 0

\sqrt{\frac{3}{2}}, \sqrt{\frac{3}{2}}

\sqrt{\frac{2}{3}}, \sqrt{\frac{2}{3}}

\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}

\frac{2}{\sqrt{3}}, \frac{2}{\sqrt{3}}

3 x^{2} - 2 \sqrt{6} x + 2 = 0

Join BYJU'S Learning Program